Chapter 8 Calculus (Q & A)

The correct option is (A).

Explanation:

By definition, a function from a set A (called the domain) to a set B (called the codomain) is a specific type of relation that satisfies two key conditions:

1. Every element in set A must be used. That is, for every element $a \in A$, there is a corresponding element $b \in B$.

2. Each element in set A is related to a unique element in set B. This means that if an element $a \in A$ is related to an element $b \in B$, it cannot be related to any other element in B.

Let's analyze the given options:

(A) A relation where each element in A is related to exactly one element in B.

This statement perfectly captures the two conditions required for a relation to be a function. It ensures that every element in the domain A has one and only one image in the codomain B.

(B) A relation where each element in B is related to exactly one element in A.

This describes a property of an injective (one-to-one) and surjective (onto) function, but it is not the fundamental definition of a function from A to B. In a general function, an element in B can be related to multiple elements in A or to no elements at all.

(C) Any subset of the Cartesian product $A \times B$.

This is the definition of a relation from A to B, not a function. A function is a special kind of relation, but not every relation is a function. For example, a relation could map one element in A to multiple elements in B, which would violate the definition of a function.

(D) A relation where each element in A is related to at least one element in B.

This statement is incomplete. While it satisfies the first condition (every element in A is used), it misses the crucial uniqueness condition. The phrase "at least one" allows for an element in A to be related to more than one element in B, which is not allowed for a function.

Therefore, option (A) is the correct and precise definition of a function from set A to set B.

The correct option is (A).

Explanation:

Let's break down the definition of a function $f: A \to B$ and analyze each option based on this definition.

A function $f$ from a set A (the domain) to a set B (the codomain) is a rule that assigns to each element $x$ in A exactly one element $y$ in B. The element $y$ is called the image of $x$ under $f$, and $x$ is a preimage of $y$.

(A) Every element in A has a unique image in B.

This statement is the precise definition of a function. It encompasses two crucial rules:

1. Every element in A is mapped: No element in the domain A is left out.
2. The mapping is unique: An element in A cannot be mapped to two or more different elements in B.

This is the fundamental property of any function.

(B) Every element in B has a unique preimage in A.

This describes a special type of function called a bijective function (or a one-to-one correspondence). It means the function is both one-to-one (injective) and onto (surjective). However, this is not true for all functions. For example, for the function $f(x) = x^2$ from $\mathbb{R} \to \mathbb{R}$, the element $4$ in B has two preimages, $2$ and $-2$. Thus, this statement is not universally true for every function.

(C) The range of $f$ is equal to B.

The range of a function is the set of all actual images, which is a subset of the codomain B. This statement says Range($f$) = B. This is the definition of a surjective (or onto) function. Not all functions are surjective. For example, for the function $f(x) = \sin(x)$ from $\mathbb{R} \to \mathbb{R}$, the range is $[-1, 1]$, which is not equal to the codomain $\mathbb{R}$.

(D) The domain of $f$ is a subset of A.

This statement is incorrect. For a function $f: A \to B$, the set A is, by definition, the entire domain of the function. Every element of A must be assigned an image. Therefore, the domain of $f$ is equal to A, not merely a subset of it.

Based on the analysis, option (A) is the only statement that correctly describes a property true for all functions.

The correct options are (C) and (D). Both of these relations fail to satisfy the definition of a function from set A to set B, but for different reasons.

Explanation:

For a relation to be a function $f: A \to B$, it must satisfy two essential conditions:

1. Every element in the domain A must be mapped. For every $a \in A$, there must exist a pair $(a, b)$ in the function's set of ordered pairs.

2. Every element in the domain A must have a unique image in B. If $(a, b_1)$ and $(a, b_2)$ are in the function's set of pairs, then it must be that $b_1 = b_2$. An input cannot map to more than one output.

Let's analyze the given options with $A = \{1, 2, 3\}$ and $B = \{p, q\}$.

(A) $\{(1, p), (2, q), (3, p)\}$: This is a function. Every element of A ({1, 2, 3}) is mapped, and each is mapped to exactly one element in B.

(B) $\{(1, p), (2, p), (3, p)\}$: This is a function (specifically, a constant function). Every element of A is mapped, and each has a unique image (even though the image is the same for all).

(C) $\{(1, p), (1, q), (2, p), (3, q)\}$: This is NOT a function. It violates the second condition. The element '1' from set A is mapped to two different elements, 'p' and 'q'. A function must assign a unique output to each input.

(D) $\{(1, p), (2, q)\}$: This is NOT a function from A to B. It violates the first condition. The element '3' from the domain set A is not mapped to any element in set B. For a function to be defined "from A", every element of A must have an image.

Since the question asks which of the options is not a function from A to B, both (C) and (D) are correct answers as they both violate a necessary condition.

The correct option is (A).

Explanation:

Given:

The function is defined as:

To Find:

The value of the function when $x = 2$, which is denoted as $f(2)$.

Solution:

To find $f(2)$, we substitute the value $x=2$ into the expression for $f(x)$.

Replace every 'x' in the function's formula with '2':

Now, we perform the arithmetic operations:

The result is 1. Comparing this with the given options, we see that it matches option (A).

The correct option is (A).

Explanation:

Let's analyze both the Assertion and the Reason.

Assertion (A): Every function is a relation.

A relation from a set A to a set B is defined as any subset of the Cartesian product $A \times B$. It is simply a set of ordered pairs $(a, b)$ where $a \in A$ and $b \in B$.

A function from set A to set B is also a set of ordered pairs, which means it is a subset of $A \times B$. However, it must adhere to specific rules. Since any function can be represented as a set of ordered pairs that is a subset of $A \times B$, it fits the definition of a relation.

Therefore, the Assertion (A) is true.

Reason (R): A function is a special type of relation where each element in the domain is paired with exactly one element in the codomain.

This statement provides the precise definition of a function. It highlights the two conditions that a relation must satisfy to be classified as a function:

1. Every element in the domain must be used ('each element ... is paired').
2. Every element in the domain must have a unique output ('...with exactly one element').

This definition correctly describes what makes a function a "special type" of relation. Therefore, the Reason (R) is also true.

Conclusion:

The Reason (R) explains exactly why the Assertion (A) is true. Assertion (A) makes a general statement (all functions are relations), and Reason (R) provides the specific definition that places functions as a sub-category within relations. It explains that a function is a relation that has additional, specific constraints. Thus, R is the correct explanation for A.

The correct option is (A).

Explanation:

Given:

The function is defined as:

To Find:

The domain of the function $f(x)$.

Solution:

The domain of a function is the set of all possible input values (x-values) for which the function is defined and produces a real number as output.

The given function is a rational function, which is a fraction. A rational function is undefined when its denominator is equal to zero, because division by zero is an undefined operation.

To find the value(s) of x for which the function is undefined, we set the denominator equal to zero:

Solving for x, we get:

This means that the function $f(x)$ is not defined at $x = 2$.

Therefore, the domain of the function consists of all real numbers except for the value 2. This can be expressed in set notation as $\mathbb{R} - \{2\}$ or as "All real numbers except 2".

Comparing this with the options provided, it matches option (A).

The correct option is (B).

Explanation:

Given:

The function is defined as:

To Find:

The domain of the function $f(x)$.

Solution:

The domain of a function is the set of all possible input values (x-values) for which the function is defined. For a function involving a square root, the expression inside the square root must be non-negative (i.e., greater than or equal to zero) to produce a real number output.

Therefore, we must satisfy the following inequality:

To solve for x, we add 3 to both sides of the inequality:

This means the function is defined for all real numbers that are greater than or equal to 3.

In interval notation, this is represented as $[3, \infty)$. The square bracket '[' next to 3 indicates that 3 is included in the domain, and the parenthesis ')' next to infinity indicates that infinity is not a number and is not included.

Comparing our result with the given options:

• (A) $(-\infty, 3]$ represents $x \le 3$.
• (B) $[3, \infty)$ represents $x \ge 3$.
• (C) $(3, \infty)$ represents $x > 3$.
• (D) All real numbers.

The correct option is (B).

The correct option is (B).

Explanation:

Given:

The function is defined as:

The domain of the function is all real numbers, $x \in \mathbb{R}$.

To Find:

The range of the function $f(x)$.

Solution:

The range of a function is the set of all possible output values (f(x)-values) that the function can produce from the given domain.

Let's analyze the output of the function $f(x) = x^2$ for different types of input values from the domain $\mathbb{R}$:

• If x is a positive number (e.g., $x=3$), the output is $f(3) = 3^2 = 9$, which is positive.
• If x is a negative number (e.g., $x=-3$), the output is $f(-3) = (-3)^2 = 9$, which is also positive. The square of any negative number is always positive.
• If x is zero ($x=0$), the output is $f(0) = 0^2 = 0$.

From this analysis, we can see that the output of the function $f(x) = x^2$ is always a non-negative number. The function can produce 0 and any positive real number as an output, but it can never produce a negative number.

Therefore, the set of all possible output values is all real numbers greater than or equal to 0. This can be written as:

In interval notation, this set is represented as $[0, \infty)$. The square bracket '[' indicates that 0 is included in the range.

Comparing this with the given options, it matches option (B).

The correct option is (D) $\mathbb{R} - \{2, -2\}$.

Explanation:

The domain of a function is the set of all possible input values (x-values) for which the function is defined.

The given function is a rational function: $f(x) = \frac{x+1}{x^2 - 4}$.

A rational function is undefined when its denominator is equal to zero, as division by zero is not allowed in mathematics.

To find the values of $x$ that are not in the domain, we must set the denominator equal to zero and solve for $x$.

Denominator: $x^2 - 4$

Setting the denominator to zero:

$x^2 - 4 = 0$

We can solve this equation by factoring the left side as a difference of squares ($a^2 - b^2 = (a-b)(a+b)$):

$(x - 2)(x + 2) = 0$

This equation holds true if either factor is zero:

$x - 2 = 0 \implies x = 2$

or

$x + 2 = 0 \implies x = -2$

So, the function $f(x)$ is undefined at $x=2$ and $x=-2$. These two values must be excluded from the domain.

The domain of the function is the set of all real numbers ($\mathbb{R}$) except for 2 and -2.

In set notation, this is written as $\mathbb{R} - \{2, -2\}$.

The correct option is (A) Both A and R are true and R is the correct explanation of A.

Explanation:

Let's analyze the Assertion and the Reason separately.

Assertion (A): The domain of $f(x) = \frac{1}{\sqrt{x}}$ is $(0, \infty)$.

To find the domain of the function $f(x) = \frac{1}{\sqrt{x}}$, we need to consider two conditions:

1. The expression inside the square root cannot be negative. For the term $\sqrt{x}$ to be defined in real numbers, the value of $x$ must be non-negative.

2. The denominator of a fraction cannot be zero. For the function $f(x)$ to be defined, the denominator $\sqrt{x}$ cannot be equal to zero.

$\sqrt{x} \neq 0$

Squaring both sides, we get:

To find the domain, we must satisfy both conditions (i) and (ii) simultaneously. Combining $x \ge 0$ and $x \neq 0$, we get $x > 0$.

In interval notation, the condition $x > 0$ is written as $(0, \infty)$.

Thus, the domain of $f(x) = \frac{1}{\sqrt{x}}$ is indeed $(0, \infty)$. So, Assertion (A) is true.

Reason (R): The expression under the square root must be non-negative, and the denominator cannot be zero.

This statement describes the exact two conditions we used to determine the domain of the function $f(x)$. Both parts of the statement are fundamental principles for defining the domain of such functions. So, Reason (R) is true.

Conclusion:

Since Reason (R) correctly states the mathematical principles that lead to the conclusion in Assertion (A), R is the correct explanation of A.

Therefore, both A and R are true, and R is the correct explanation of A.

The correct option is (C) All non-negative integers.

Explanation:

The domain of a function refers to the set of all possible input values (in this case, $x$) for which the function is defined and makes sense in the given context.

In this problem, the variable $x$ represents the number of items produced.

We need to consider the nature of this variable:

1. Discrete vs. Continuous: The number of items produced must be a whole quantity. A company cannot produce a fraction of an item (e.g., 2.5 items). Therefore, $x$ must be an integer.

2. Non-negativity: The problem explicitly states that the number of items produced "cannot be negative". This means $x$ must be greater than or equal to zero ($x \ge 0$). A company can choose to produce zero items.

Combining these two conditions, $x$ must be an integer that is also greater than or equal to zero. The set of numbers that satisfies this is {0, 1, 2, 3, ...}.

This set is known as the set of non-negative integers (or whole numbers).

Let's analyze the given options:

(A) All real numbers: Incorrect. This would include negative numbers and fractions, which are not possible for the number of items.

(B) All positive real numbers: Incorrect. This excludes the possibility of producing 0 items and includes fractions.

(C) All non-negative integers: Correct. This set {0, 1, 2, ...} perfectly describes the possible number of items that can be produced.

(D) All integers: Incorrect. This includes negative numbers, which are not a valid number of items to produce.

Therefore, in the context of this problem, the domain for the cost function $C(x)$ is the set of all non-negative integers.

The correct option is (C) $\{10, 11, \dots, 50\}$.

Explanation:

The question asks for the "range of items produced" when the company produces between 10 and 50 items, inclusive. In this context, "range" refers to the set of all possible values for the number of items, $x$.

1. Nature of the Variable: As established in the case study, $x$ represents the number of items. Since items are discrete, countable objects, $x$ must be an integer. It is not possible to produce a fractional number of items (e.g., 25.5 items).

2. Bounds and Inclusivity: The problem states the production is "between 10 and 50 items (inclusive)". This means that the smallest number of items is 10 and the largest is 50, and both these values are included.

Combining these points, the set of possible values for $x$ is the set of all integers from 10 to 50, inclusive. This is the set {10, 11, 12, ..., 49, 50}.

Now let's analyze the options:

(A) $(10, 50)$: This is open interval notation. It represents all real numbers strictly between 10 and 50. It is incorrect because it excludes the endpoints 10 and 50 and includes non-integer values.

(B) $[10, 50]$: This is closed interval notation. It represents all real numbers between 10 and 50, including the endpoints. It is incorrect because it includes non-integer values like 10.5, 20.1, etc.

(C) $\{10, 11, \dots, 50\}$: This is set notation that explicitly lists the integers from 10 to 50. This accurately represents the discrete and inclusive nature of the number of items. This is the correct representation.

(D) $(10, 50]$: This is half-open interval notation. It represents all real numbers greater than 10 and less than or equal to 50. It is incorrect because it excludes the lower endpoint 10 and includes non-integer values.

Therefore, the correct way to describe the set of possible items produced is with set notation for integers.

The correct option is (B) One-one function (Injective).

Explanation:

The definition of a one-one function, also known as an injective function, is a function where every distinct element in the domain is mapped to a distinct element in the codomain.

Mathematically, a function $f: A \to B$ is one-one if for any two elements $x_1$ and $x_2$ in the domain $A$:

If $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$.

This is precisely what is described in the question.

Let's look at why the other options are incorrect:

(A) Onto function (Surjective): An onto function is one where every element in the codomain has at least one corresponding element in the domain. It ensures the entire codomain is "covered" by the function's outputs, but it does not guarantee that different inputs have different outputs. For example, $f(x) = x(x-1)(x+1)$ is an onto function from $\mathbb{R}$ to $\mathbb{R}$, but it is not one-one ($f(1)=f(0)=f(-1)=0$).

(C) Many-one function: This is the opposite of a one-one function. A many-one function maps two or more distinct elements of the domain to the same element in the codomain. For example, $f(x) = x^2$ is a many-one function because $f(-2) = 4$ and $f(2) = 4$.

(D) Constant function: A constant function maps all elements of the domain to a single element in the codomain (e.g., $f(x) = c$ for all $x$). This is an extreme example of a many-one function.

Therefore, the only type of function that guarantees distinct inputs map to distinct outputs is a one-one (injective) function.

The correct option is (B) Every element in B has at least one preimage in A.

Explanation:

An onto function, also known as a surjective function, is a function $f: A \to B$ where every element in the codomain (set B) is an image of at least one element from the domain (set A). Another way to state this is that the range of the function is equal to its codomain.

Let's analyze the options based on this definition:

(A) Every element in A has a unique image in B.

This is a fundamental property of any function, not specifically an onto function. It means that for each input, there is exactly one output.

(B) Every element in B has at least one preimage in A.

A "preimage" of an element $y \in B$ is an element $x \in A$ such that $f(x) = y$. This statement says that for every possible output value in the codomain B, there is at least one input value in the domain A that produces it. This is the precise definition of an onto function.

(C) The range of $f$ is a proper subset of B.

This is the definition of a function that is not onto. A proper subset means the range is contained within the codomain but is not equal to it, implying there are elements in the codomain B that are not mapped to by any element in A.

(D) Every element in A is mapped to the same element in B.

This describes a constant function. A constant function is only onto if the codomain B consists of a single element. This is not the general definition of an onto function.

Therefore, option (B) correctly defines an onto (surjective) function.

The correct option is (C) Many-one but not onto.

Explanation:

The given function is $f: \mathbb{R} \to \mathbb{R}$, defined by $f(x) = x^2$.

To determine the nature of the function, we need to check if it is one-one (injective) and onto (surjective).

Check for One-one (Injective) Property:

A function is one-one if different elements in the domain have different images in the codomain. That is, if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

Let's consider two distinct elements from the domain $\mathbb{R}$, for example, $x_1 = -2$ and $x_2 = 2$.

$f(-2) = (-2)^2 = 4$

$f(2) = (2)^2 = 4$

Here, we have $f(-2) = f(2)$, but $-2 \neq 2$. Since two different elements in the domain (-2 and 2) map to the same element in the codomain (4), the function is not one-one. It is a many-one function.

Check for Onto (Surjective) Property:

A function is onto if its range is equal to its codomain. The codomain is given as $\mathbb{R}$ (the set of all real numbers).

The range of the function is the set of all possible output values. For $f(x) = x^2$, the output is always non-negative, since the square of any real number is greater than or equal to zero.

So, the range of $f(x)$ is $[0, \infty)$, which is the set of all non-negative real numbers.

The range $[0, \infty)$ is a proper subset of the codomain $\mathbb{R}$ (i.e., Range $\neq$ Codomain). For example, there is no real number $x$ in the domain such that $f(x) = x^2 = -1$, even though -1 is an element of the codomain $\mathbb{R}$.

Since the range is not equal to the codomain, the function is not onto.

Conclusion:

The function $f(x) = x^2$ is both many-one and not onto. Therefore, option (C) is the correct choice.

The correct option is (C) Both one-one and onto (Bijective).

Explanation:

The identity function on a set A is given by $I_A: A \to A$, where $I_A(x) = x$ for every element $x \in A$. This means the function maps every element to itself.

To determine its properties, we check for both one-one (injective) and onto (surjective) conditions.

Check for One-one (Injective) Property:

A function is one-one if $I_A(x_1) = I_A(x_2)$ implies $x_1 = x_2$.

Let's assume $I_A(x_1) = I_A(x_2)$ for any two elements $x_1, x_2 \in A$.

By the definition of the identity function, $I_A(x_1) = x_1$ and $I_A(x_2) = x_2$.

So, the assumption $I_A(x_1) = I_A(x_2)$ directly leads to $x_1 = x_2$.

Since this condition holds for all elements, the identity function is one-one.

Check for Onto (Surjective) Property:

A function is onto if its range is equal to its codomain. In this case, the codomain is the set A.

The range is the set of all possible output values. Since $I_A(x) = x$ for every $x \in A$, the set of all outputs is simply the set A itself.

Range of $I_A = \{I_A(x) \mid x \in A\} = \{x \mid x \in A\} = A$.

Since the Range (A) is equal to the Codomain (A), the identity function is onto.

Conclusion:

Since the identity function is both one-one and onto, it is a bijective function. Therefore, option (C) is the correct choice.

The correct option is (C) Constant function.

Explanation:

The function is given by $f(x) = c$, where $c$ is a constant. This is the definition of a constant function, as it maps every input value $x$ from the domain to the same single, constant value $c$ in the codomain.

Let's analyze why the other options are incorrect in this general case:

(A) One-one function: A function is one-one if different inputs produce different outputs. For a constant function, let's take two different inputs, say $x_1 = 1$ and $x_2 = 2$.

$f(1) = c$

$f(2) = c$

Since $f(1) = f(2)$ but $1 \neq 2$, the function is not one-one. It is a many-one function.

(B) Onto function: A function is onto if its range is equal to its codomain. The codomain is given as $\mathbb{R}$ (all real numbers). The range of $f(x) = c$ is the set of all possible outputs, which is just the single element set $\{c\}$. Since the range $\{c\}$ is not equal to the codomain $\mathbb{R}$, the function is not onto.

(D) Bijective function: A function is bijective if it is both one-one and onto. Since the constant function $f(x)=c$ is neither one-one nor onto, it cannot be bijective.

Therefore, the most accurate and defining characteristic of the function $f(x) = c$ is that it is a constant function.

The correct option is (A) Both A and R are true and R is the correct explanation of A.

Explanation:

To evaluate this question, we must first analyze the truthfulness of the Assertion and the Reason individually, and then determine if the Reason correctly explains the Assertion.

Analysis of Assertion (A):

Assertion (A) states that the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^3$ is a bijective function. A function is bijective if it is both one-one (injective) and onto (surjective).

1. Checking for One-one (Injective):

A function is one-one if $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

Let $x_1, x_2 \in \mathbb{R}$ such that $f(x_1) = f(x_2)$.

$x_1^3 = x_2^3$

Taking the cube root of both sides, we get:

$\sqrt[3]{x_1^3} = \sqrt[3]{x_2^3}$

$x_1 = x_2$

Since this is true for all real numbers (unlike with $x^2$), the function is one-one.

2. Checking for Onto (Surjective):

A function is onto if for every element $y$ in the codomain ($\mathbb{R}$), there exists an element $x$ in the domain ($\mathbb{R}$) such that $f(x) = y$.

Let $y \in \mathbb{R}$. We need to find an $x$ such that $f(x) = y$.

$x^3 = y$

Solving for $x$, we get $x = \sqrt[3]{y}$ or $x = y^{1/3}$.

For any real number $y$ (whether positive, negative, or zero), its real cube root $x = y^{1/3}$ exists and is also a real number. For example, if $y=8, x=2$; if $y=-27, x=-3$. Therefore, the range of the function is $\mathbb{R}$, which is equal to the codomain. The function is onto.

Since the function is both one-one and onto, it is bijective. Thus, Assertion (A) is true.

Analysis of Reason (R):

Reason (R) states that for $f(x) = x^3$, every real number $y$ has a unique real cube root $x = y^{1/3}$. This statement has two components:

• "every real number $y$ has a ... real cube root": This implies that for any $y$ in the codomain, a pre-image $x$ exists. This is the condition for the function to be onto.
• "every real number $y$ has a unique real cube root": This implies that each $y$ in the range comes from exactly one $x$. This is the condition for the function to be one-one.

Both components of the statement are mathematically correct. Thus, Reason (R) is true.

Conclusion:

Reason (R) states that the function has a unique output for every input and covers the entire codomain, which is precisely the definition of a bijective function. It provides the exact justification for why the function is both one-one and onto. Therefore, R is the correct explanation of A.

The correct option is (B) One-one and onto function.

Explanation:

Let's analyze the properties of the function $f: A \to B$ based on the given information.

The domain is the set of students, A. The size of the domain is $|A| = 10$.

The codomain is the set of roll numbers, B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. The size of the codomain is $|B| = 10$.

1. Checking for One-one (Injective) Property:

The problem states that "each student is assigned a unique roll number". This means that if we take two different students, student₁ and student₂, then their assigned roll numbers will also be different, i.e., $f(\text{student₁}) \neq f(\text{student₂})$.

This is the definition of a one-one function. Different inputs map to different outputs.

2. Checking for Onto (Surjective) Property:

A function is onto if its range is equal to its codomain. The codomain is the set of all roll numbers, B = {1, 2, ..., 10}.

We have 10 students and 10 roll numbers. Since each student gets a unique roll number, all 10 roll numbers from the set B must be used. There are no unassigned roll numbers left.

Therefore, the range of the function (the set of assigned roll numbers) is {1, 2, ..., 10}, which is exactly the same as the codomain B.

Since the range equals the codomain, the function is onto.

Conclusion:

Since the function is both one-one and onto, it is a bijective function. Option (B) correctly describes this.

The correct option is (B) (i)-(c), (ii)-(a), (iii)-(b).

Explanation:

Let's analyze each function type and match it with its correct definition from the list.

(i) One-one function (Injective):

A function is one-one if every distinct element in the domain maps to a distinct element in the codomain. The formal way to test this is to assume that two outputs are equal, $f(x_1) = f(x_2)$, and then show that this must imply that the inputs were also equal, $x_1 = x_2$. This matches property (c).

So, (i) matches with (c).

(ii) Onto function (Surjective):

A function is onto if its range is equal to its codomain. This means that for every element $y$ in the codomain, there is at least one element $x$ in the domain that maps to it, such that $f(x) = y$. This is exactly what property (a) states.

So, (ii) matches with (a).

(iii) Bijective function:

A function is bijective by definition if it is both one-one and onto. This directly corresponds to property (b).

So, (iii) matches with (b).

Therefore, the correct matching is:

• (i) $\to$ (c)
• (ii) $\to$ (a)
• (iii) $\to$ (b)

This corresponds to option (B).

The correct option is (B) y-axis.

Explanation:

In the graphical representation of a function in a Cartesian coordinate system, the two axes have specific roles:

1. The domain of the function, which is the set of all possible input values, is represented along the horizontal axis, commonly known as the x-axis.

2. The range of the function, which is the set of all possible output values (or function values), is represented along the vertical axis, commonly known as the y-axis.

To find the range of a function from its graph, one would look at all the vertical values that the graph covers. This is equivalent to projecting the entire graph onto the y-axis.

Let's analyze the options:

(A) x-axis: This axis represents the domain of the function, not the range.

(B) y-axis: This axis represents the output values, which constitute the range of the function. This is correct.

(C) origin: The origin is the single point (0, 0) and does not represent the set of all possible values for the range.

(D) horizontal axis: This is another name for the x-axis, which represents the domain.

Therefore, the range corresponds to the set of all possible values on the y-axis.

The correct option is (B) vertical line.

Explanation:

The test used to determine if a graph represents a function is called the Vertical Line Test.

The definition of a function requires that for every input value (x-value) in the domain, there must be exactly one output value (y-value). Graphically, this means that any vertical line drawn on the coordinate plane can intersect the graph of the function at most once.

If a vertical line intersects the graph at more than one point, it implies that there is a single x-value corresponding to multiple y-values, which violates the definition of a function.

Let's consider the other options:

(A) horizontal line test: This test is used to determine if a function is one-to-one (injective). If any horizontal line intersects the graph more than once, the function is not one-to-one.

(C) diagonal line test: This is not a standard test used in function analysis.

(D) curved line test: This is not a standard test used in function analysis.

Therefore, the correct answer is the vertical line test.

The correct option is (B) The graph does not represent a function.

Explanation:

This question directly relates to the Vertical Line Test, which is a graphical method to determine if a curve in the plane represents the graph of a function.

The definition of a function requires that for every input value (x-coordinate) in its domain, there must be exactly one corresponding output value (y-coordinate).

When you draw a vertical line on a graph, that line represents a single, constant x-value. If this vertical line intersects the graph at more than one point, it means that for that single x-value, there are multiple y-values. This violates the fundamental definition of a function.

Let's analyze the options:

(A) The graph represents a function: This is incorrect. The condition described is the exact reason why a graph would fail to be a function.

(B) The graph does not represent a function: This is the correct conclusion from the Vertical Line Test.

(C) The function is one-one: This is incorrect. First, the graph does not represent a function. Second, the one-one property is tested using the horizontal line test, not the vertical line test.

(D) The function is onto: This is incorrect. The concept of "onto" (or surjective) relates the function's range to its codomain and cannot be determined by the vertical line test. Furthermore, if the graph doesn't represent a function, discussing its properties like being onto is irrelevant.

The correct option is (B) y-axis.

Explanation:

A graph is symmetric about the y-axis if it represents an even function. An even function is a function $f(x)$ that satisfies the property $f(-x) = f(x)$ for all values of $x$ in its domain.

Let's test the given function $f(x) = |x|$ to see if it is an even function.

We need to evaluate $f(-x)$:

$f(-x) = |-x|$

By the definition of the absolute value, the absolute value of a number and its negative are the same. For example, $|-3| = 3$ and $|3| = 3$. Therefore, $|-x| = |x|$.

So, we have:

$f(-x) = |-x| = |x| = f(x)$

Since $f(-x) = f(x)$, the function $f(x) = |x|$ is an even function. The graphs of all even functions are symmetric with respect to the y-axis.

Visually, the graph of $f(x) = |x|$ is a "V" shape with its vertex at the origin. The right side of the V (where $x > 0$) is a mirror image of the left side of the V (where $x < 0$) across the y-axis.

Therefore, the graph of the function $f(x) = |x|$ is symmetric about the y-axis.

The correct option is (B) A straight line.

Explanation:

The function given is $f(x) = mx + c$. This is the standard form of a linear function.

By definition, the graph of any linear function is always a straight line.

Let's break down the components of the function:

• $m$ represents the slope or gradient of the line. It determines the steepness and direction of the line. The condition $m \neq 0$ means the line is not horizontal.
• $c$ represents the y-intercept, which is the point where the line crosses the y-axis.

Now, let's examine the options:

(A) A curve: While a straight line is technically a type of curve, this term is usually used to describe non-linear paths. The most specific and accurate description is "a straight line".

(B) A straight line: This is the correct and precise graphical representation of a linear function.

(C) A parabola: A parabola is the graph of a quadratic function, which has the form $f(x) = ax^2 + bx + c$ where $a \neq 0$.

(D) A circle: A circle is represented by an equation like $(x-h)^2 + (y-k)^2 = r^2$. A circle is not a function because it fails the vertical line test.

Therefore, the graph of a linear function is a straight line.

The correct option is (C) $f(x) = |x|$.

Explanation:

The question provides a piecewise function and asks which standard function it represents.

The given function is:

$f(x) = \begin{cases} x & , & x \geq 0 \\ -x & , & x < 0 \end{cases}$

This is the precise mathematical definition of the absolute value function, which is denoted as $f(x) = |x|$.

Let's verify this definition:

• If $x$ is non-negative (e.g., $x=5$ or $x=0$), the absolute value is the number itself. For $x=5$, $|5| = 5$. The piecewise function uses the first rule ($x \ge 0$), giving $f(5) = 5$. This matches.
• If $x$ is negative (e.g., $x=-5$), the absolute value is the positive version of that number. For $x=-5$, $|-5| = 5$. The piecewise function uses the second rule ($x < 0$), giving $f(-5) = -(-5) = 5$. This also matches.

Since the piecewise function is the definition of the absolute value function, their graphs are identical.

Let's look at the other options:

(A) $f(x) = x$: This is a straight line passing through the origin with a slope of 1. It only matches the first part of the piecewise function.

(B) $f(x) = x^2$: This is a parabola (a U-shaped curve) with its vertex at the origin.

(D) $f(x) = \sqrt{x}$: This is a curve that only exists for $x \geq 0$.

The graph of the given piecewise function is a characteristic V-shape, which is the graph of $f(x) = |x|$.

The correct option is (A) Both A and R are true and R is the correct explanation of A.

Explanation:

To answer this question, we need to evaluate both the Assertion and the Reason and then determine if the Reason correctly explains the Assertion.

Analysis of Assertion (A):

Assertion (A) states that the graph of a circle is not the graph of a function. The definition of a function requires that for every input $x$ in the domain, there is exactly one output $y$. For a circle centered at the origin, its equation is $x^2 + y^2 = r^2$. If we pick an $x$-value such that $-r < x < r$, we get two corresponding $y$-values: $y = \sqrt{r^2 - x^2}$ and $y = -\sqrt{r^2 - x^2}$. Since one input value leads to two output values, a circle does not satisfy the definition of a function. Therefore, Assertion (A) is true.

Analysis of Reason (R):

Reason (R) states that a vertical line can intersect a circle at two points, violating the vertical line test. The Vertical Line Test is the graphical method for determining if a curve is the graph of a function. If any vertical line can be drawn that intersects the graph more than once, the graph does not represent a function. For any circle, a vertical line drawn through its interior (not at the tangent points) will intersect the circle at two points. This is a correct statement. Therefore, Reason (R) is true.

Conclusion:

The reason that a circle is not a function (Assertion A) is precisely because it fails the Vertical Line Test (Reason R). The Reason provides the direct and correct justification for the Assertion. Thus, both A and R are true, and R is the correct explanation of A.

The correct option is (D) $[1, \infty)$.

Explanation:

The range of a function is the set of all possible output values (y-values) that the function can produce. When looking at a graph, the range is the set of all y-values covered by the curve.

The problem describes the graph as a parabola opening upwards with its vertex at the point (0, 1).

For a parabola that opens upwards, its vertex represents the minimum point on the graph. This means the lowest y-value the function can have is the y-coordinate of the vertex.

In this case, the minimum y-value is 1.

Since the parabola extends infinitely upwards from this minimum point, the function's output values (the range) include 1 and all real numbers greater than 1.

In interval notation, the set of all numbers greater than or equal to 1 is written as $[1, \infty)$. The square bracket "[" is used to indicate that the endpoint 1 is included in the set.

The correct option is (B) No.

Explanation:

To determine if a function is one-one (or injective) from its graph, we use the Horizontal Line Test.

The Horizontal Line Test states that a function is one-one if and only if no horizontal line intersects its graph at more than one point.

The graph described is a parabola opening upwards. This shape is symmetrical about its vertical axis of symmetry (in this case, the y-axis).

If you draw any horizontal line above the vertex of the parabola (which is at y=1), the line will intersect the graph at two distinct points. For example, a line like $y = 5$ would cut the parabola on both its left and right sides.

This means that there are two different input values (x-values) that produce the same output value (y-value). For instance, for the function $f(x)=x^2+1$ which matches the description, we have:

$f(-2) = (-2)^2 + 1 = 5$

$f(2) = (2)^2 + 1 = 5$

Since two different inputs (-2 and 2) lead to the same output (5), the function is not one-one. It is a many-to-one function.

Therefore, the function represented by the graph is not one-one.

The correct option is (C) The value of $f(x)$ gets arbitrarily close to $L$ as $x$ gets arbitrarily close to $a$ (but not equal to $a$).

Explanation:

The concept of a limit in calculus is about the value a function "approaches" as the input approaches a certain point. It does not depend on the actual value of the function at that point.

Let's analyze the options:

(A) $f(a) = L$: This statement says that the function's value *at* $x=a$ is $L$. While this is true for functions that are continuous at $a$, it is not the definition of a limit. The limit can exist even if $f(a)$ is different from $L$ or if $f(a)$ is not defined at all.

(B) The value of $f(x)$ is exactly $L$ at $x=a$: This is another way of phrasing option (A) and is incorrect for the same reason.

(C) The value of $f(x)$ gets arbitrarily close to $L$ as $x$ gets arbitrarily close to $a$ (but not equal to $a$): This is the correct informal definition of a limit. It captures the essential idea that we are interested in the behavior of the function *near* the point $x=a$, not necessarily *at* the point $x=a$. The phrase "but not equal to $a$" is crucial.

(D) $f(x)$ is defined at $x=a$ and $f(a)=L$: This describes a function that is continuous at $x=a$. The existence of a limit is a necessary condition for continuity, but it is not sufficient. The limit itself is a more fundamental concept that doesn't require the function to be defined or continuous at the point in question.

Therefore, option (C) provides the most accurate and general definition of a limit.

The correct option is (C) 7.

Explanation:

We are asked to evaluate the limit: $\lim\limits_{x \to 2} (3x + 1)$.

The function is $f(x) = 3x + 1$. This is a linear function, which is a type of polynomial function.

For polynomial functions, the limit as $x$ approaches any real number $a$ can be found by direct substitution of $a$ into the function. This is because polynomial functions are continuous everywhere.

So, we can substitute $x = 2$ into the expression:

$\lim\limits_{x \to 2} (3x + 1) = 3(2) + 1$

$\implies \lim\limits_{x \to 2} (3x + 1) = 6 + 1$

$\implies \lim\limits_{x \to 2} (3x + 1) = 7$

Therefore, the limit of the function as $x$ approaches 2 is 7.

The correct option is (B) 6.

Explanation:

We need to evaluate the limit: $\lim\limits_{x \to 3} \frac{x^2 - 9}{x - 3}$.

First, if we try direct substitution by plugging in $x = 3$, we get:

$\frac{3^2 - 9}{3 - 3} = \frac{9 - 9}{3 - 3} = \frac{0}{0}$

This is an indeterminate form, which suggests that we should try to simplify the expression algebraically.

The numerator, $x^2 - 9$, is a difference of squares, which can be factored as $a^2 - b^2 = (a-b)(a+b)$.

$x^2 - 9 = (x - 3)(x + 3)$

Now, we can substitute the factored form back into the limit:

$\lim\limits_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3}$

Since the limit concerns the value as $x$ approaches 3 (but is not equal to 3), the term $(x - 3)$ in the numerator and denominator is non-zero and can be cancelled out.

$\lim\limits_{x \to 3} \frac{\cancel{(x - 3)}(x + 3)}{\cancel{x - 3}} = \lim\limits_{x \to 3} (x + 3)$

Now we can evaluate the simplified limit by direct substitution:

$\lim\limits_{x \to 3} (x + 3) = 3 + 3 = 6$

Therefore, the limit is 6.

Alternate Solution (Using L'Hôpital's Rule):

Since direct substitution leads to the indeterminate form $\frac{0}{0}$, we can apply L'Hôpital's Rule. This rule allows us to take the derivative of the numerator and the denominator separately and then find the limit.

Let $f(x) = x^2 - 9$ and $g(x) = x - 3$.

The derivative of the numerator is $f'(x) = 2x$.

The derivative of the denominator is $g'(x) = 1$.

Applying the rule:

$\lim\limits_{x \to 3} \frac{f(x)}{g(x)} = \lim\limits_{x \to 3} \frac{f'(x)}{g'(x)} = \lim\limits_{x \to 3} \frac{2x}{1}$

Now, by direct substitution:

$\lim\limits_{x \to 3} 2x = 2(3) = 6$

Both methods give the same result.

The correct option is (C) $\lim\limits_{x \to a} f(x) = f(a)$.

Explanation:

The formal definition of continuity of a function $f(x)$ at a point $x=a$ combines three distinct conditions into a single, comprehensive statement.

The three conditions are:

1. The function must be defined at $x=a$. That is, $f(a)$ must exist.
2. The limit of the function as $x$ approaches $a$ must exist. That is, $\lim\limits_{x \to a} f(x)$ exists.
3. The value of the limit must be equal to the value of the function at that point.

Let's analyze the given options in light of these conditions:

(A) $\lim\limits_{x \to a} f(x)$ exists: This is a necessary condition, but not sufficient. A function could have a limit at a point but still not be continuous if the function's value at that point is different or undefined (a "hole" in the graph).

(B) $f(a)$ is defined: This is also a necessary condition, but not sufficient. A function can be defined at a point, but the limit might not exist (a "jump" in the graph).

(C) $\lim\limits_{x \to a} f(x) = f(a)$: This single statement is the complete definition of continuity at a point. It implicitly requires that both sides of the equation are well-defined:

• For the left side, $\lim\limits_{x \to a} f(x)$, to have a value, the limit must exist (satisfying condition 2).
• For the right side, $f(a)$, to have a value, the function must be defined at $a$ (satisfying condition 1).
• The equality itself is the final condition (condition 3).

(D) All of the above conditions are met: While this statement is true for a continuous function, it is redundant. Option (C) already implies that (A) and (B) must be true. In mathematics, the most concise and complete definition is preferred, which is option (C).

The correct option is (C) 0.

Explanation:

The problem states that the function $f(x) = |x|$ is continuous at the point $x=0$.

By the definition of continuity at a point $x=a$, we know that three conditions must be met, which are summarized by the single equation:

$\lim\limits_{x \to a} f(x) = f(a)$

In this specific case, $a=0$ and the function is $f(x) = |x|$. So, for continuity at $x=0$, we have:

$\lim\limits_{x \to 0} |x| = f(0)$

We can find the value of $f(0)$ by substituting $x=0$ into the function:

$f(0) = |0| = 0$

Since the limit must be equal to the function's value at the point of continuity, we can conclude:

$\lim\limits_{x \to 0} |x| = 0$

Alternate Method (Direct Evaluation of the Limit):

We can also evaluate the limit by checking the left-hand limit (LHL) and the right-hand limit (RHL).

1. Right-Hand Limit (RHL): As $x$ approaches 0 from the positive side ($x \to 0^+$), $x$ is positive, so $|x| = x$.

$\lim\limits_{x \to 0^+} |x| = \lim\limits_{x \to 0^+} x = 0$

2. Left-Hand Limit (LHL): As $x$ approaches 0 from the negative side ($x \to 0^-$), $x$ is negative, so $|x| = -x$.

$\lim\limits_{x \to 0^-} |x| = \lim\limits_{x \to 0^-} (-x) = -0 = 0$

Since LHL = RHL = 0, the limit exists and is equal to 0.

The correct option is (B) No.

Explanation:

For a function $f(x)$ to be continuous at a point $x=a$, it must satisfy three conditions:

1. $f(a)$ must be defined.
2. $\lim\limits_{x \to a} f(x)$ must exist.
3. $\lim\limits_{x \to a} f(x) = f(a)$.

Let's check the first condition for the function $f(x) = \frac{1}{x-2}$ at the point $a=2$.

We evaluate the function at $x=2$:

$f(2) = \frac{1}{2-2} = \frac{1}{0}$

Division by zero is undefined. Therefore, the first condition, that $f(a)$ must be defined, is not met.

Since the function is not defined at $x=2$, it cannot be continuous at that point. There is a non-removable (infinite) discontinuity at $x=2$.

Option (D) is incorrect because the limit does not exist either; as $x$ approaches 2 from the right, $f(x)$ approaches $+\infty$, and as $x$ approaches 2 from the left, $f(x)$ approaches $-\infty$.

The correct option is (C) A is false but R is true.

Explanation:

Let's analyze both the Assertion and the Reason.

Analysis of Assertion (A):

The assertion states that the existence of a limit is a sufficient condition for continuity. This is false. The existence of the limit is only one of the three conditions required for continuity. A function can have a limit at a point but still fail to be continuous there.

Counterexample: Consider a function with a removable discontinuity (a "hole").

$f(x) = \frac{x^2-4}{x-2}$

The limit as $x$ approaches 2 exists:

$\lim\limits_{x \to 2} \frac{(x-2)(x+2)}{x-2} = \lim\limits_{x \to 2} (x+2) = 4$

However, the function $f(x)$ is not defined at $x=2$. Since one of the conditions for continuity ($f(2)$ must be defined) fails, the function is not continuous at $x=2$. Thus, the assertion is false.

Analysis of Reason (R):

The reason states the three conditions for continuity at a point $x=a$:

1. $\lim\limits_{x \to a} f(x)$ must exist.
2. $f(a)$ must be defined.
3. $\lim\limits_{x \to a} f(x) = f(a)$.

Conclusion:

Since Assertion (A) is false and Reason (R) is true, the correct option is (C).

The correct option is (C) 3.

Explanation:

The given function is a piecewise function:

$f(x) = \begin{cases} x+2 & , & x < 1 \\ kx & , & x \geq 1 \end{cases}$

For a function to be continuous at a point $x=a$, the left-hand limit (LHL) must equal the right-hand limit (RHL), and this common value must be equal to the function's value at that point. In short:

$\lim\limits_{x \to a^-} f(x) = \lim\limits_{x \to a^+} f(x) = f(a)$

In this problem, we need to ensure continuity at the point where the function definition changes, which is $x=1$.

1. Find the Left-Hand Limit (LHL):

The LHL is the limit as $x$ approaches 1 from values less than 1 ($x < 1$). For this case, we use the expression $f(x) = x+2$.

$\text{LHL} = \lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^-} (x+2)$

By substituting $x=1$, we get:

$\text{LHL} = 1 + 2 = 3$

2. Find the Right-Hand Limit (RHL):

The RHL is the limit as $x$ approaches 1 from values greater than or equal to 1 ($x \geq 1$). For this case, we use the expression $f(x) = kx$.

$\text{RHL} = \lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1^+} (kx)$

By substituting $x=1$, we get:

$\text{RHL} = k(1) = k$

3. Equate the limits:

For the function to be continuous at $x=1$, the LHL must equal the RHL.

$\text{LHL} = \text{RHL}$

$3 = k$

Therefore, the value of $k$ must be 3 for the function to be continuous at $x=1$.

The correct option is (C) 3.

Explanation:

The function is given by the piecewise definition:

$f(x) = \begin{cases} x+2 & , & x < 1 \\ kx & , & x \geq 1 \end{cases}$

The left-hand limit (LHL) as $x$ approaches 1, denoted by $\lim\limits_{x \to 1^-} f(x)$, considers the values of the function as $x$ gets close to 1 from the left side, i.e., for values of $x$ that are less than 1 ($x < 1$).

According to the function's definition, for the condition $x < 1$, the expression for $f(x)$ is $x+2$.

Therefore, to find the left-hand limit, we use this expression:

$\text{LHL} = \lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^-} (x+2)$

We can find this limit by substituting $x=1$ into the expression:

$\text{LHL} = 1 + 2 = 3$

Thus, the left-hand limit of $f(x)$ as $x$ approaches 1 is 3.

The correct option is (B) The slope of the tangent line to the graph at $x=a$.

Explanation:

The instantaneous rate of change of a function at a specific point is one of the fundamental concepts in differential calculus. It is defined as the limit of the average rate of change as the interval shrinks to zero. This value is also known as the derivative of the function at that point, denoted by $f'(a)$.

Let's analyze the options from a geometric perspective:

(A) The slope of the secant line through two points on the graph: A secant line passes through two points on a curve. Its slope represents the average rate of change of the function over the interval between those two points. It is not the instantaneous rate of change.

(B) The slope of the tangent line to the graph at $x=a$: A tangent line touches the graph at a single point ($x=a$) and represents the direction of the curve at that exact point. The slope of this tangent line is precisely the definition of the instantaneous rate of change (and the derivative) of the function at that point. This is the correct answer.

(C) The average rate of change over an interval: This is represented by the slope of the secant line (as in option A), not the tangent line. It gives a rate of change over a duration, not at a single instant.

(D) The value of the function at $x=a$: This is $f(a)$, which represents the position or value of the function at that point, not its rate of change.

The correct option is (C) derivative.

Explanation:

The expression given in the question is one of the most fundamental definitions in calculus.

The expression $\frac{f(a+h) - f(a)}{h}$ represents the slope of the secant line between the points $(a, f(a))$ and $(a+h, f(a+h))$. This is the average rate of change of the function over the small interval of length $h$.

Taking the limit as $h$ approaches 0, $\lim\limits_{h \to 0}$, transforms this average rate of change into an instantaneous rate of change at the single point $x=a$.

This limit, $\lim\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$, is the formal definition of the derivative of the function $f(x)$ at the point $x=a$, often denoted as $f'(a)$.

Let's look at the options:

• (A) integral: The integral is related to finding the area under a curve and is the inverse operation of differentiation.
• (B) value: The value of the function at $x=a$ is simply $f(a)$, not the rate of change.
• (C) derivative: This is the correct term for the given limit definition.
• (D) limit: While the expression is a limit, its specific form defines the derivative. "Derivative" is the precise and specific name for this concept.

The correct option is (B) Instantaneous rate of change.

Explanation:

The derivative of a function with respect to a variable measures the instantaneous rate of change of that function with respect to that variable.

In this context:

• The function is the position, $s(t)$.
• The variable is time, $t$.

The derivative of the position function, $s'(t)$ or $\frac{ds}{dt}$, gives the rate at which the position is changing at a specific moment in time. By definition, the instantaneous rate of change of position is instantaneous velocity.

Therefore, calculating instantaneous velocity from a position function is a direct physical application of the mathematical concept of an instantaneous rate of change.

Let's consider the other options:

• Average rate of change / Average velocity: This is the change in position over a time interval ($\frac{\Delta s}{\Delta t}$), not at a single instant. It is represented by the slope of a secant line, not the derivative.
• Total displacement: This is the overall change in position, $\Delta s = s(t_{final}) - s(t_{initial})$. It is a change in quantity, not a rate of change.

The correct option is (C) $f'(x)$ or $\frac{dy}{dx}$.

Explanation:

The derivative of a function measures its instantaneous rate of change. There are several standard notations used to represent the derivative.

Let's analyze the options:

(A) $\int f(x) dx$: This is the notation for the indefinite integral of $f(x)$ with respect to $x$. Integration is the inverse operation of differentiation.

(B) $f(x)$: This denotes the function itself, not its derivative.

(C) $f'(x)$ or $\frac{dy}{dx}$: Both of these are standard notations for the derivative.

• $f'(x)$ is known as Lagrange's notation. The prime symbol (') indicates differentiation with respect to the function's variable.
• $\frac{dy}{dx}$ (assuming $y=f(x)$) is known as Leibniz's notation. It explicitly shows that the function $y$ is being differentiated with respect to the variable $x$.

(D) $\lim\limits_{x \to a} f(x)$: This denotes the limit of the function $f(x)$ as $x$ approaches a specific value $a$. While the derivative is defined using a limit, this notation represents the limit of the function itself, not the derivative.

The correct option is (C) differentiation.

Explanation:

The term differentiation is the mathematical process used to calculate the derivative of a function. The derivative itself represents the instantaneous rate of change of the function.

Let's look at the other options:

• (A) integration: This is the process of finding the integral of a function, which is the inverse operation of differentiation. It is often used to find the area under a curve.
• (B) limits: Limits are a foundational concept in calculus that are used to define both derivatives and integrals, but "limits" is not the name for the process of finding a derivative.
• (D) evaluation: This is a general term that means to find the value of a mathematical expression. It is not specific to the process of finding a derivative.

The correct option is (C) $3x^2$.

Explanation:

The power rule for differentiation is a fundamental rule used to find the derivative of functions of the form $f(x) = x^n$. The rule states:

$\frac{d}{dx}(x^n) = nx^{n-1}$

In words, to find the derivative, you bring the exponent ($n$) down as a coefficient and then subtract 1 from the original exponent.

For the given function, $f(x) = x^3$, the exponent is $n=3$.

Applying the power rule:

$f'(x) = \frac{d}{dx}(x^3) = 3 \cdot x^{3-1}$

$\implies f'(x) = 3x^2$

Therefore, the derivative of $f(x) = x^3$ is $3x^2$.

Option (D) $x^4/4$ is the integral of $x^3$, not the derivative.

The correct option is (A) $10x - 7$.

Explanation:

To find the derivative of the polynomial function $f(x) = 5x^2 - 7x + 2$, we need to differentiate it term by term using the sum/difference rule, the constant multiple rule, and the power rule.

The derivative is given by:

$f'(x) = \frac{d}{dx}(5x^2 - 7x + 2)$

$f'(x) = \frac{d}{dx}(5x^2) - \frac{d}{dx}(7x) + \frac{d}{dx}(2)$

Now let's differentiate each term:

1. Derivative of $5x^2$: Using the power rule ($\frac{d}{dx}(x^n) = nx^{n-1}$), the derivative of $x^2$ is $2x^{2-1} = 2x$. We multiply this by the constant 5:

$\frac{d}{dx}(5x^2) = 5 \cdot (2x) = 10x$

2. Derivative of $7x$: The term $x$ is $x^1$. The derivative of $x^1$ is $1x^{1-1} = 1x^0 = 1$. We multiply this by the constant 7:

$\frac{d}{dx}(7x) = 7 \cdot (1) = 7$

3. Derivative of 2: The derivative of any constant is 0.

$\frac{d}{dx}(2) = 0$

Combining the results:

$f'(x) = 10x - 7 + 0$

$f'(x) = 10x - 7$

The correct option is (D) 0.

Explanation:

The derivative of a function measures its instantaneous rate of change. We are asked to find the derivative of a constant function, $f(x) = c$ (in this case, $f(x)=10$).

There are a few ways to understand why the derivative is 0:

1. Conceptual Understanding (Rate of Change):

A constant function, by definition, does not change its value. For any change in $x$, the change in $f(x)$ is always zero. Since the rate of change is zero, the derivative is 0.

2. Graphical Interpretation (Slope):

The graph of a constant function like $f(x) = 10$ is a horizontal line at $y=10$. The derivative of a function at a point is the slope of the tangent line at that point. For a horizontal line, the slope is always 0. Thus, the derivative is 0.

3. Using the Formal Rule:

The rule for differentiating a constant is one of the most basic rules in calculus:

$\frac{d}{dx}(c) = 0$, where $c$ is any constant.

In this case, $c=10$, so $\frac{d}{dx}(10) = 0$.

The correct option is (A) $f'(x) + g'(x)$.

Explanation:

This question refers to a fundamental property of derivatives known as the Sum Rule for Differentiation.

The Sum Rule states that the derivative of a sum of two differentiable functions is the sum of their individual derivatives.

If $h(x) = f(x) + g(x)$, then the derivative of $h(x)$ is given by:

$h'(x) = f'(x) + g'(x)$

Using the notation from the question, this is expressed as:

$(f+g)'(x) = f'(x) + g'(x)$

This rule allows us to differentiate complex functions by breaking them down into simpler parts, differentiating each part, and then adding the results. It is the reason we can differentiate a polynomial term by term.

The correct option is (A) Both A and R are true and R is the correct explanation of A.

Explanation:

Let's analyze the Assertion and Reason.

Assertion (A): The derivative of $f(x) = x^n$ is $nx^{n-1}$.

This is a statement of one of the most fundamental rules of differential calculus. The formula is correct. Therefore, Assertion (A) is true.

Reason (R): This is the power rule of differentiation.

The rule stated in the assertion, $\frac{d}{dx}(x^n) = nx^{n-1}$, is indeed known as the power rule. The name correctly identifies the rule. Therefore, Reason (R) is true.

Conclusion:

The Reason (R) correctly provides the name for the mathematical rule stated in the Assertion (A). In this context, identifying the rule by its proper name serves as a correct explanation or classification of the assertion. Therefore, both A and R are true, and R is the correct explanation of A.

The correct option is (B) $6(2x+1)^2$.

Explanation:

To find the derivative of a composite function like $f(x) = (2x+1)^3$, we use the Chain Rule.

The Chain Rule states that if a function $y$ can be written as a composition of two functions, say $y = g(u)$ where $u = h(x)$, then the derivative of $y$ with respect to $x$ is the derivative of the outer function multiplied by the derivative of the inner function:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

For the given problem, $f(x) = (2x+1)^3$, we can set:

• The inner function: $u = 2x+1$
• The outer function: $y = u^3$

Now, we find the derivatives of these two parts:

1. Differentiate the outer function ($y = u^3$) with respect to $u$:

Using the power rule, $\frac{dy}{du} = 3u^2$.

2. Differentiate the inner function ($u = 2x+1$) with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(2x+1) = 2$.

3. Apply the Chain Rule by multiplying the two derivatives:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = (3u^2) \cdot (2) = 6u^2$.

4. Substitute the expression for $u$ back into the result:

Since $u = 2x+1$, we replace $u$ in the derivative:

$\frac{dy}{dx} = 6(2x+1)^2$.

Therefore, the derivative of $f(x) = (2x+1)^3$ is $6(2x+1)^2$.

The correct option is (B) $\frac{dy}{du} \times \frac{du}{dx}$.

Explanation:

This question asks for the formal statement of the Chain Rule in Leibniz notation, which is the fundamental rule for differentiating composite functions.

A composite function is created when one function is nested inside another. In this case, $y$ is a function of $u$, and $u$ is a function of $x$. We can think of $y$ as the "outer function" and $u$ as the "inner function".

The Chain Rule states that the derivative of the composite function with respect to $x$ is the product of the derivative of the outer function (with respect to its variable) and the derivative of the inner function (with respect to its variable).

The formula for the Chain Rule is:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

This corresponds exactly to option (B).

The correct option is (C) $\frac{x}{\sqrt{x^2+1}}$.

Explanation:

We are asked to find the derivative of the function $f(x) = \sqrt{x^2 + 1}$.

This is a composite function, so we must use the Chain Rule. First, we can rewrite the function using a rational exponent:

$f(x) = (x^2 + 1)^{1/2}$

The Chain Rule states that for a composite function $y = g(h(x))$, the derivative is $y' = g'(h(x)) \cdot h'(x)$. In words, this is the derivative of the outer function (with the inner function left inside) multiplied by the derivative of the inner function.

Let's identify the parts:

• The outer function is the power function, $(\cdot)^{1/2}$.
• The inner function is the expression inside the parentheses, $x^2 + 1$.

Now, let's apply the rule in steps:

1. Differentiate the outer function:

Apply the power rule to $(\cdot)^{1/2}$. The derivative is $\frac{1}{2}(\cdot)^{-1/2}$. We keep the inner function, $x^2 + 1$, inside this derivative:

$\frac{1}{2}(x^2 + 1)^{-1/2} = \frac{1}{2\sqrt{x^2+1}}$

2. Differentiate the inner function:

The derivative of the inner function, $x^2 + 1$, is:

$\frac{d}{dx}(x^2 + 1) = 2x + 0 = 2x$

3. Multiply the results from Step 1 and Step 2:

According to the chain rule, we multiply the derivative of the outer function by the derivative of the inner function.

$f'(x) = \left( \frac{1}{2\sqrt{x^2+1}} \right) \cdot (2x)$

$f'(x) = \frac{2x}{2\sqrt{x^2+1}}$

We can simplify this by cancelling the 2 in the numerator and the denominator:

$f'(x) = \frac{\cancel{2}x}{\cancel{2}\sqrt{x^2+1}} = \frac{x}{\sqrt{x^2+1}}$

Therefore, the derivative is $\frac{x}{\sqrt{x^2+1}}$.

The correct option is (C) $4(x^2 - 3x + 5)^3 (2x - 3)$.

Explanation:

The function $y = (x^2 - 3x + 5)^4$ is a composite function, so we must use the Chain Rule to find its derivative, $\frac{dy}{dx}$.

The Chain Rule in this context can be thought of as:

$\frac{dy}{dx} = (\text{derivative of the outer function}) \times (\text{derivative of the inner function})$

Let's identify the parts of our function:

• The outer function is the power function, $(\cdot)^4$.
• The inner function is the polynomial inside the parentheses, $x^2 - 3x + 5$.

Now, we differentiate step-by-step:

1. Differentiate the outer function:

Apply the power rule to $(\cdot)^4$. The derivative is $4(\cdot)^3$. We keep the inner function, $(x^2 - 3x + 5)$, unchanged inside:

$4(x^2 - 3x + 5)^3$

2. Differentiate the inner function:

The derivative of the inner function, $x^2 - 3x + 5$, is:

$\frac{d}{dx}(x^2 - 3x + 5) = 2x - 3 + 0 = 2x - 3$

3. Multiply the results from Step 1 and Step 2:

According to the Chain Rule, we multiply the two results together to get the final derivative:

$\frac{dy}{dx} = \underbrace{4(x^2 - 3x + 5)^3}_{\text{Derivative of outer}} \cdot \underbrace{(2x - 3)}_{\text{Derivative of inner}}$

So, $\frac{dy}{dx} = 4(x^2 - 3x + 5)^3 (2x - 3)$.

The correct option is (A) $-2(x+1)^{-3}$.

Explanation:

To find the derivative of $f(x) = \frac{1}{(x+1)^2}$, we should first rewrite the function using a negative exponent. This makes it easier to apply the power rule and chain rule.

$f(x) = (x+1)^{-2}$

This is a composite function where:

• The outer function is $(\cdot)^{-2}$.
• The inner function is $x+1$.

We will use the Chain Rule to find the derivative, $f'(x)$.

1. Differentiate the outer function:

Apply the power rule to $(\cdot)^{-2}$. The derivative is $-2(\cdot)^{-2-1} = -2(\cdot)^{-3}$. We keep the inner function, $(x+1)$, inside:

$-2(x+1)^{-3}$

2. Differentiate the inner function:

The derivative of the inner function, $x+1$, is:

$\frac{d}{dx}(x+1) = 1 + 0 = 1$

3. Multiply the results from Step 1 and Step 2:

According to the Chain Rule, we multiply the two results:

$f'(x) = (-2(x+1)^{-3}) \cdot (1)$

$f'(x) = -2(x+1)^{-3}$

This can also be written in fraction form as $\frac{-2}{(x+1)^3}$. The answer provided in the options uses negative exponents, so option (A) is the correct form.

The correct option is (A) Both A and R are true and R is the correct explanation of A.

Explanation:

Let's analyze both the Assertion and the Reason to determine their validity and relationship.

Analysis of Assertion (A):

The assertion provides a calculation for the derivative of $f(x) = (x^2+5)^2$ using the chain rule. Let's verify it.

The function is a composition with:

• Outer function: $(\cdot)^2$
• Inner function: $x^2+5$

According to the chain rule, the derivative is the derivative of the outer function (with the inner function intact) multiplied by the derivative of the inner function.

1. Derivative of the outer part: $2(x^2+5)^1 = 2(x^2+5)$

2. Derivative of the inner part: $\frac{d}{dx}(x^2+5) = 2x$

3. Multiplying them: $f'(x) = 2(x^2+5) \times (2x) = 4x(x^2+5)$

The calculation shown in the assertion is correct. Therefore, Assertion (A) is true.

Analysis of Reason (R):

The reason states that the chain rule is applied to composite functions. This is the fundamental definition of the purpose of the chain rule in calculus. Therefore, Reason (R) is true.

Conclusion:

Assertion (A) gives a specific example of applying the chain rule. Reason (R) states the general principle of when to use the chain rule. Since the function in (A) is a composite function, the reason (R) correctly explains why the chain rule is the appropriate method to use. Thus, both are true, and R is the correct explanation for A.

The correct option is (A) $4\pi r^2$.

Explanation:

This question asks for the derivative of the volume of a sphere, $V$, with respect to its radius, $r$. This value, $\frac{dV}{dr}$, represents the instantaneous rate of change of the volume as the radius changes.

The formula for the volume of a sphere is given as:

$V = \frac{4}{3}\pi r^3$

To find $\frac{dV}{dr}$, we need to differentiate the function $V$ with respect to the variable $r$. In this formula, $\frac{4}{3}$ and $\pi$ are constants.

We use the power rule for differentiation, which states that $\frac{d}{dx}(x^n) = nx^{n-1}$. Here, our variable is $r$ and the function is $r^3$.

$\frac{dV}{dr} = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right)$

We treat the constant part, $\frac{4}{3}\pi$, as a coefficient:

$\frac{dV}{dr} = \frac{4}{3}\pi \cdot \frac{d}{dr}(r^3)$

Now, we apply the power rule to $r^3$:

$\frac{d}{dr}(r^3) = 3r^{3-1} = 3r^2$

Substitute this back into our expression:

$\frac{dV}{dr} = \frac{4}{3}\pi \cdot (3r^2)$

We can simplify this by cancelling the 3 in the denominator with the 3 in the numerator:

$\frac{dV}{dr} = \frac{4}{\cancel{3}}\pi \cdot (\cancel{3}r^2)$

$\frac{dV}{dr} = 4\pi r^2$

Interestingly, the result, $4\pi r^2$, is the formula for the surface area of the sphere. This means the rate of change of a sphere's volume with respect to its radius is equal to its surface area.

The correct option is (B) $800\pi \text{ cm}^3/\text{s}$.

Explanation:

We are given the rate at which the radius is changing ($\frac{dr}{dt}$) and are asked to find the rate at which the volume is changing ($\frac{dV}{dt}$) at a specific instant. This is a related rates problem that uses the chain rule.

The relationship between the volume $V$ and the radius $r$ of a sphere is given by:

$V = \frac{4}{3}\pi r^3$

To relate their rates of change with respect to time $t$, we differentiate both sides of this equation with respect to $t$. Since $V$ and $r$ are both functions of time, we must use the Chain Rule on the right side.

$\frac{d}{dt}(V) = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)$

The Chain Rule states that $\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}$.

So, our equation becomes:

$\frac{dV}{dt} = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right) \cdot \frac{dr}{dt}$

From the previous question, we know the derivative of the volume with respect to the radius is:

$\frac{dV}{dr} = 4\pi r^2$

Substituting this into our rate equation, we get:

$\frac{dV}{dt} = (4\pi r^2) \cdot \frac{dr}{dt}$

Now, we use the specific values given for the instant in question:

• The radius, $r = 10$ cm
• The rate of change of the radius, $\frac{dr}{dt} = 2$ cm/s

Substitute these values into the equation:

$\frac{dV}{dt} = (4\pi (10)^2) \cdot (2)$

$\frac{dV}{dt} = (4\pi \cdot 100) \cdot 2$

$\frac{dV}{dt} = 400\pi \cdot 2$

$\frac{dV}{dt} = 800\pi$

The units for the rate of change of volume are cubic centimeters per second (cm³/s).

Therefore, the volume is increasing at a rate of $800\pi \text{ cm}^3/\text{s}$ at that instant.

The correct option is (A) $(-2, 2)$.

Explanation:

To find the domain of the function $f(x) = \frac{1}{\sqrt{4-x^2}}$, we must consider two conditions that need to be satisfied simultaneously.

Condition 1: The expression inside the square root must be non-negative.

The argument of a square root cannot be negative for the function to be defined in the set of real numbers.

Condition 2: The denominator of the fraction cannot be zero.

The entire denominator, $\sqrt{4-x^2}$, cannot be equal to zero.

$\sqrt{4-x^2} \neq 0$

Squaring both sides gives:

Combining the conditions:

We need to satisfy both $4 - x^2 \ge 0$ and $4 - x^2 \neq 0$. Together, these conditions simplify to a single strict inequality:

$4 - x^2 > 0$

Now, we solve this inequality:

$4 > x^2$

This is equivalent to:

$x^2 < 4$

Taking the square root of both sides, we get:

$|x| < 2$

This absolute value inequality means that $x$ must be between -2 and 2, not including the endpoints.

$-2 < x < 2$

In interval notation, this set of values is represented by the open interval $(-2, 2)$.

Therefore, the domain of the function is $(-2, 2)$.

The correct option is (B) $\{5\}$.

Explanation:

The range of a function is the set of all possible output values (y-values) that the function can produce.

The given function is $f(x) = 5$. This is a constant function.

By its definition, a constant function maps every single input value from its domain to one single, constant output value. In this case, no matter what value of $x$ we choose as input, the output will always be 5.

For example:

• $f(0) = 5$
• $f(-10) = 5$
• $f(100) = 5$

Since the only possible output value is 5, the range is the set containing just this single element.

In set notation, this is written as $\{5\}$.

The correct option is (B) $4x + 9$.

Explanation:

The notation $f(f(x))$ represents the composition of the function $f(x)$ with itself. To find this, we take the entire expression for $f(x)$ and use it as the input for the function $f(x)$ again.

The given function is:

$f(x) = 2x + 3$

To find $f(f(x))$, we substitute the expression for $f(x)$, which is $(2x+3)$, in place of every $x$ in the original function definition:

$f(f(x)) = f(2x+3)$

Now we apply the rule $f(\text{input}) = 2(\text{input}) + 3$ with our new input:

$f(f(x)) = 2(2x+3) + 3$

Next, we simplify the expression by distributing the 2 and then combining the constant terms:

$f(f(x)) = (2 \cdot 2x) + (2 \cdot 3) + 3$

$f(f(x)) = 4x + 6 + 3$

$f(f(x)) = 4x + 9$

Therefore, the composition $f(f(x))$ is equal to $4x+9$.

The correct option is (B) 0.

Explanation:

We are asked to evaluate the limit: $\lim\limits_{x \to 0} \frac{x^2}{x}$.

If we attempt to use direct substitution by plugging in $x=0$, we get:

$\frac{0^2}{0} = \frac{0}{0}$

This is an indeterminate form, which means we must simplify the expression before evaluating the limit.

We can simplify the fraction $\frac{x^2}{x}$ for values of $x$ where $x \neq 0$.

$\frac{x^2}{x} = x$

Since the limit is concerned with values of $x$ approaching 0 (but not equal to 0), we can replace the original expression with its simplified form:

$\lim\limits_{x \to 0} \frac{x^2}{x} = \lim\limits_{x \to 0} x$

Now, we can evaluate this simplified limit by direct substitution:

$\lim\limits_{x \to 0} x = 0$

Therefore, the value of the limit is 0.

The correct option is (A) True.

Explanation:

This is a fundamental theorem in calculus. If a function is differentiable at a point, it is guaranteed to be continuous at that same point.

Why is this true?

The definition of the derivative at a point $x=a$ is:

$f'(a) = \lim\limits_{x \to a} \frac{f(x) - f(a)}{x - a}$

For a function to be differentiable, this limit must exist and be a finite number.

To prove continuity, we need to show that $\lim\limits_{x \to a} f(x) = f(a)$, which is the same as showing that $\lim\limits_{x \to a} [f(x) - f(a)] = 0$.

We can write $f(x) - f(a)$ as:

$f(x) - f(a) = \left( \frac{f(x) - f(a)}{x - a} \right) \cdot (x - a)$

Now, taking the limit of both sides as $x \to a$:

$\lim\limits_{x \to a} [f(x) - f(a)] = \lim\limits_{x \to a} \left[ \left( \frac{f(x) - f(a)}{x - a} \right) \cdot (x - a) \right]$

$\lim\limits_{x \to a} [f(x) - f(a)] = \left( \lim\limits_{x \to a} \frac{f(x) - f(a)}{x - a} \right) \cdot \left( \lim\limits_{x \to a} (x - a) \right)$

From the definition of the derivative, the first term is $f'(a)$, and the second term is $a-a=0$.

$\lim\limits_{x \to a} [f(x) - f(a)] = f'(a) \cdot 0 = 0$

This shows that $\lim\limits_{x \to a} f(x) = f(a)$, which is the definition of continuity.

It is important to note that the converse is not true: a function can be continuous at a point without being differentiable there. A common example is the absolute value function $f(x) = |x|$ at $x=0$. It is continuous but has a sharp corner, so it is not differentiable at that point.

The correct option is (A) $28x^3 - 6x^2 + 5$.

Explanation:

To find the derivative of the polynomial function $f(x) = 7x^4 - 2x^3 + 5x - 1$, we need to differentiate it term by term. We will use the following rules of differentiation:

• Power Rule: $\frac{d}{dx}(x^n) = nx^{n-1}$
• Constant Multiple Rule: $\frac{d}{dx}(c \cdot g(x)) = c \cdot g'(x)$
• Sum/Difference Rule: The derivative of a sum/difference is the sum/difference of the derivatives.
• Constant Rule: The derivative of a constant is 0.

Let's differentiate each term of the polynomial:

1. Derivative of $7x^4$:

Using the constant multiple rule and the power rule:

$\frac{d}{dx}(7x^4) = 7 \cdot \frac{d}{dx}(x^4) = 7 \cdot (4x^{4-1}) = 28x^3$

2. Derivative of $-2x^3$:

$\frac{d}{dx}(-2x^3) = -2 \cdot \frac{d}{dx}(x^3) = -2 \cdot (3x^{3-1}) = -6x^2$

3. Derivative of $5x$:

Remember that $x$ is $x^1$.

$\frac{d}{dx}(5x) = 5 \cdot \frac{d}{dx}(x^1) = 5 \cdot (1x^{1-1}) = 5 \cdot (x^0) = 5 \cdot 1 = 5$

4. Derivative of $-1$:

Using the constant rule:

$\frac{d}{dx}(-1) = 0$

Now, we combine the derivatives of each term:

$f'(x) = 28x^3 - 6x^2 + 5 - 0$

$f'(x) = 28x^3 - 6x^2 + 5$

Therefore, the derivative is $28x^3 - 6x^2 + 5$.

The correct option is (C) $-5(5x - 2)^{-2}$.

Explanation:

The function $y = (5x - 2)^{-1}$ is a composite function, which means we must use the Chain Rule to find its derivative, $\frac{dy}{dx}$.

The Chain Rule states that the derivative of a composite function is the derivative of the outer function multiplied by the derivative of the inner function.

Let's identify the parts of our function:

• The outer function is the power function, $(\cdot)^{-1}$.
• The inner function is the expression inside the parentheses, $5x - 2$.

Now, we differentiate step-by-step:

1. Differentiate the outer function:

Using the power rule, $\frac{d}{du}(u^n) = nu^{n-1}$, the derivative of the outer function is $-1(\cdot)^{-1-1} = -1(\cdot)^{-2}$. We keep the inner function, $(5x - 2)$, unchanged inside this expression:

$-1(5x - 2)^{-2}$

2. Differentiate the inner function:

The derivative of the inner function, $5x - 2$, with respect to $x$ is:

$\frac{d}{dx}(5x - 2) = 5 - 0 = 5$

3. Multiply the results from Step 1 and Step 2:

According to the Chain Rule, we multiply the two results together:

$\frac{dy}{dx} = [-1(5x - 2)^{-2}] \cdot [5]$

$\frac{dy}{dx} = -5(5x - 2)^{-2}$

The correct option is (C) 0.

Explanation:

The derivative of a function at a point represents the slope of the tangent line to the function's graph at that point. It also represents the instantaneous rate of change of the function.

The function given is $f(x) = c$, a constant function. Its graph is a horizontal line.

By definition, a horizontal line has a slope of zero at every point. Since the derivative is equal to the slope of the tangent line, and the tangent to a horizontal line is the line itself, the derivative must be 0.

From the perspective of rate of change, a constant function does not change its value as $x$ changes. Therefore, its rate of change is 0.

The formal rule for differentiating a constant is:

$\frac{d}{dx}(c) = 0$

Thus, the derivative of the constant function is 0.

The correct option is (A) $\{-1, 1\}$.

Explanation:

The range of a function is the set of all possible output values (y-values) it can produce.

The function is given by $f(x) = \frac{|x|}{x}$. To determine its range, we need to analyze its behavior for different values of $x$.

First, note that the function is undefined at $x=0$ because it would lead to division by zero. So, 0 is not in the domain of the function.

Now, let's consider the two possible cases for $x \neq 0$:

Case 1: $x$ is positive ($x > 0$)

If $x$ is a positive number, then by definition, $|x| = x$.

Substituting this into the function:

$f(x) = \frac{x}{x} = 1$

So, for any positive input, the output is always 1.

Case 2: $x$ is negative ($x < 0$)

If $x$ is a negative number, then by definition, $|x| = -x$.

Substituting this into the function:

$f(x) = \frac{-x}{x} = -1$

So, for any negative input, the output is always -1.

Since these are the only two possibilities for the input $x$ (it cannot be zero), the only two possible output values for the function are 1 and -1.

Therefore, the range of the function is the set containing just these two values: $\{-1, 1\}$.

The correct option is (B) $\lim\limits_{x \to 1} \frac{|x-1|}{x-1}$.

Explanation:

A limit exists at a point if and only if the left-hand limit (LHL) is equal to the right-hand limit (RHL) at that point. Let's analyze each option.

(A) $\lim\limits_{x \to 0} x^2$:

This is a polynomial function, which is continuous everywhere. We can find the limit by direct substitution.

$\lim\limits_{x \to 0} x^2 = (0)^2 = 0$. The limit exists.

(B) $\lim\limits_{x \to 1} \frac{|x-1|}{x-1}$:

Because of the absolute value, we must check the LHL and RHL.

• Right-Hand Limit (RHL): As $x$ approaches 1 from the right ($x \to 1^+$), $x > 1$, so $(x-1)$ is positive. Thus, $|x-1| = x-1$.

$\lim\limits_{x \to 1^+} \frac{|x-1|}{x-1} = \lim\limits_{x \to 1^+} \frac{x-1}{x-1} = \lim\limits_{x \to 1^+} 1 = 1$

• Left-Hand Limit (LHL): As $x$ approaches 1 from the left ($x \to 1^-$), $x < 1$, so $(x-1)$ is negative. Thus, $|x-1| = -(x-1)$.

$\lim\limits_{x \to 1^-} \frac{|x-1|}{x-1} = \lim\limits_{x \to 1^-} \frac{-(x-1)}{x-1} = \lim\limits_{x \to 1^-} -1 = -1$

Since the LHL ($-1$) is not equal to the RHL ($1$), the limit does not exist.

(C) $\lim\limits_{x \to 2} \frac{x^2 - 4}{x-2}$:

Direct substitution gives the indeterminate form $\frac{0}{0}$. We can simplify by factoring the numerator.

$\lim\limits_{x \to 2} \frac{(x-2)(x+2)}{x-2} = \lim\limits_{x \to 2} (x+2) = 2+2 = 4$. The limit exists.

(D) $\lim\limits_{x \to 5} (x+5)$:

This is a linear function, so we can use direct substitution.

$\lim\limits_{x \to 5} (x+5) = 5+5 = 10$. The limit exists.

Therefore, the only limit that does not exist is the one in option (B).

The correct option is (C) $5(x^3 + 2x)^4 (3x^2 + 2)$.

Explanation:

The function $f(x) = (x^3 + 2x)^5$ is a composite function, so we must use the Chain Rule to find its derivative, $f'(x)$.

The Chain Rule states that the derivative of a composite function is the derivative of the outer function (with the inner function left intact) multiplied by the derivative of the inner function.

Let's identify the parts of the function:

• The outer function is the power function, $(\cdot)^5$.
• The inner function is the polynomial inside the parentheses, $x^3 + 2x$.

Now, we differentiate step-by-step:

1. Differentiate the outer function:

Using the power rule, the derivative of $(\cdot)^5$ is $5(\cdot)^4$. We keep the inner function, $(x^3 + 2x)$, unchanged inside this expression:

$5(x^3 + 2x)^4$

2. Differentiate the inner function:

The derivative of the inner function, $x^3 + 2x$, with respect to $x$ is:

$\frac{d}{dx}(x^3 + 2x) = 3x^2 + 2$

3. Multiply the results from Step 1 and Step 2:

According to the Chain Rule, we multiply the two results together to get the final derivative:

$f'(x) = \underbrace{5(x^3 + 2x)^4}_{\text{Derivative of outer}} \cdot \underbrace{(3x^2 + 2)}_{\text{Derivative of inner}}$

So, the derivative is $5(x^3 + 2x)^4 (3x^2 + 2)$.

The correct option is (D) $a \neq 0$.

Explanation:

The function $f(x) = ax + b$ represents a straight line with slope $a$ and y-intercept $b$. For this function to be both one-one and onto (bijective) from $\mathbb{R}$ to $\mathbb{R}$, we need to analyze the role of the slope $a$.

Case 1: $a = 0$

If $a=0$, the function becomes $f(x) = b$. This is a constant function, and its graph is a horizontal line.

• One-one test: A horizontal line fails the horizontal line test because multiple (in fact, all) x-values map to the same y-value, $b$. Therefore, it is not one-one.
• Onto test: The range of this function is the single set $\{b\}$. The codomain is $\mathbb{R}$. Since the range is not equal to the codomain, the function is not onto.

So, we must have $a \neq 0$.

Case 2: $a \neq 0$

If $a \neq 0$, the graph is a non-horizontal straight line.

• One-one test: Any non-horizontal line (whether its slope is positive or negative) will always pass the horizontal line test. Every horizontal line will intersect the graph at exactly one point. Therefore, the function is one-one.
• Onto test: A non-horizontal line extends infinitely in both the positive and negative y-directions. Its range is the set of all real numbers, $(-\infty, \infty)$ or $\mathbb{R}$. This is equal to the codomain. Therefore, the function is onto.

Conclusion:

For the linear function $f(x) = ax+b$ to be both one-one and onto, the only condition that must be met is that its slope, $a$, is not equal to zero.

The correct option is (B) $(1, 5]$.

Explanation:

To find the domain of the function $f(x) = \frac{1}{\sqrt{x-1}} + \sqrt{5-x}$, we must find the set of all real numbers $x$ for which the function is defined. This requires finding the intersection of the domains of the two separate terms in the function.

Let's consider the two parts of the function:

Part 1: The term $\frac{1}{\sqrt{x-1}}$

For this term to be defined, two conditions must be met:

1. The expression inside the square root must be non-negative: $x-1 \ge 0$.

2. The denominator cannot be zero, which means $\sqrt{x-1} \neq 0$, or $x-1 \neq 0$.

Combining these two conditions, we get a single strict inequality:

Part 2: The term $\sqrt{5-x}$

For this term to be defined, the expression inside the square root must be non-negative:

Finding the Overall Domain:

The domain of the entire function $f(x)$ is the intersection of the values of $x$ that satisfy both condition (i) and condition (ii).

We need $x$ to be greater than 1 AND less than or equal to 5.

$x > 1$ and $x \le 5$

This can be written as the compound inequality:

$1 < x \le 5$

In interval notation, this corresponds to the interval $(1, 5]$. The parenthesis on the left indicates that 1 is not included, and the square bracket on the right indicates that 5 is included.

The correct option is (A) True.

Explanation:

This statement is a fundamental concept in calculus regarding the relationship between the graph of a function and its differentiability.

Differentiability and Smoothness:

A function is differentiable at a point if its graph is "smooth" and does not have any abrupt changes in direction at that point. Geometrically, this means that there is a single, well-defined, non-vertical tangent line to the curve at that point. The slope of this tangent line is the value of the derivative.

Sharp Corners and Cusps:

1. A sharp corner occurs when the slope of the graph changes abruptly. A classic example is the function $f(x) = |x|$ at the point $x=0$. The slope to the left of 0 is -1, and the slope to the right is +1. At the corner itself, there is no single, unique tangent line. The left-hand derivative and the right-hand derivative are not equal, so the limit that defines the derivative does not exist.

2. A cusp is a point on the graph where the tangent line becomes vertical. At a vertical tangent, the slope is infinite, which means the derivative is undefined. An example is the function $f(x) = x^{2/3}$ at $x=0$.

In both cases—sharp corners and cusps—the function is not differentiable at that point because a unique, finite slope does not exist. Therefore, the statement is true.

The correct option is (A) $-5x^{-6}$.

Explanation:

To find the derivative of the function $f(x) = \frac{1}{x^5}$, the first step is to rewrite the function using a negative exponent. This allows us to use the power rule for differentiation.

$f(x) = x^{-5}$

The power rule for differentiation states that for any function of the form $g(x) = x^n$, its derivative is given by:

$\frac{d}{dx}(x^n) = nx^{n-1}$

For our function, $f(x) = x^{-5}$, the exponent is $n = -5$.

Now, we apply the power rule:

$f'(x) = \frac{d}{dx}(x^{-5}) = (-5) \cdot x^{-5 - 1}$

Simplifying the exponent:

$f'(x) = -5x^{-6}$

This result can also be written in fraction form as $\frac{-5}{x^6}$. The options are given in terms of negative exponents, so $-5x^{-6}$ is the correct form.

The correct option is (B) $[2, \infty)$.

Explanation:

The range of a function is the set of all possible output values (y-values or $f(x)$ values) it can produce.

The function is given by $f(x) = x^2 + 2$. The domain is all real numbers ($x \in \mathbb{R}$).

To find the range, we can analyze the components of the function:

1. The term $x^2$ is the square of a real number. The square of any real number is always non-negative (greater than or equal to zero).

2. The function adds 2 to this non-negative value. We can add 2 to both sides of the inequality to find the possible values for the entire function:

$x^2 + 2 \ge 0 + 2$

$x^2 + 2 \ge 2$

Since $f(x) = x^2 + 2$, this means:

$f(x) \ge 2$

This shows that the minimum possible value of the function is 2 (which occurs when $x=0$), and the function can take on any value greater than or equal to 2.

In interval notation, the set of all real numbers greater than or equal to 2 is written as $[2, \infty)$. The square bracket "[" indicates that the endpoint 2 is included in the range.

Graphical Method:

The graph of $y=x^2$ is a parabola opening upwards with its vertex at the origin (0,0). The graph of $y = x^2 + 2$ is the same parabola shifted vertically upwards by 2 units. Therefore, its vertex is at the point (0, 2). Since the parabola opens upwards, this vertex is the minimum point. The lowest y-value on the graph is 2, and it extends infinitely upwards. Thus, the range is $[2, \infty)$.

The correct option is (D) does not exist.

Explanation:

This question asks about the fundamental condition for the existence of a limit.

The limit of a function $f(x)$ as $x$ approaches a point $a$, denoted as $\lim\limits_{x \to a} f(x)$, exists if and only if two conditions are met:

1. The left-hand limit (LHL), $\lim\limits_{x \to a^-} f(x)$, exists.
2. The right-hand limit (RHL), $\lim\limits_{x \to a^+} f(x)$, exists.
3. The left-hand limit is equal to the right-hand limit (LHL = RHL).

If these conditions are met, the value of the limit is this common value.

The problem states that the left-hand limit is not equal to the right-hand limit:

$\lim\limits_{x \to a^-} f(x) \neq \lim\limits_{x \to a^+} f(x)$

This directly violates the third condition required for the existence of the overall limit. When the function approaches different values from the left and the right of the point $a$, there is no single value that the function is approaching. This is often seen as a "jump" in the graph of the function.

Therefore, if the left-hand and right-hand limits are not equal, the two-sided limit $\lim\limits_{x \to a} f(x)$ does not exist.

The correct option is (B) $f(x) = |x|$.

Explanation:

A function is continuous for all real numbers if it is defined and continuous at every point on the number line. We can analyze each option:

(A) $f(x) = \frac{1}{x}$:

This is a rational function. It is not defined at $x=0$ because this would cause division by zero. Since the function is not defined at $x=0$, it is not continuous for all real numbers. It has an infinite discontinuity at $x=0$.

(B) $f(x) = |x|$:

This is the absolute value function. Its graph is a "V" shape with the vertex at the origin. The graph has no breaks, holes, or jumps. It is continuous for all real numbers. We can prove this by checking the point $x=0$:

• $\lim\limits_{x \to 0^-} |x| = \lim\limits_{x \to 0^-} (-x) = 0$
• $\lim\limits_{x \to 0^+} |x| = \lim\limits_{x \to 0^+} (x) = 0$
• $f(0) = |0| = 0$

(C) $f(x) = \sqrt{x}$:

This is the square root function. For the function to be defined in the real number system, the expression inside the square root must be non-negative. Therefore, the domain of this function is $x \ge 0$. Since it is not defined for negative real numbers, it is not continuous for all real numbers.

(D) $f(x) = \begin{cases} 0 & , & x \leq 0 \\ 1 & , & x > 0 \end{cases}$:

This is a piecewise function (the Heaviside step function). We must check for continuity at the point where the definition changes, $x=0$.

• The left-hand limit is $\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} 0 = 0$.
• The right-hand limit is $\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} 1 = 1$.

The correct option is (C) $\frac{2x - 5}{2\sqrt{x^2 - 5x}}$. (Note: Option (D) is also a correct representation of the same answer).

Explanation:

The function is given by $f(x) = (x^2 - 5x)^{1/2}$, which can also be written as $f(x) = \sqrt{x^2 - 5x}$.

To find the derivative of this composite function, we must use the Chain Rule. The rule states that the derivative of a composite function is the derivative of the outer function multiplied by the derivative of the inner function.

Let's identify the parts of our function:

• The outer function is the power function, $(\cdot)^{1/2}$.
• The inner function is the polynomial inside the parentheses, $x^2 - 5x$.

We differentiate step-by-step:

1. Differentiate the outer function:

Using the power rule, the derivative of $(\cdot)^{1/2}$ is $\frac{1}{2}(\cdot)^{-1/2}$. We keep the inner function, $(x^2 - 5x)$, inside this expression:

$\frac{1}{2}(x^2 - 5x)^{-1/2}$

2. Differentiate the inner function:

The derivative of the inner function, $x^2 - 5x$, with respect to $x$ is:

$\frac{d}{dx}(x^2 - 5x) = 2x - 5$

3. Multiply the results from Step 1 and Step 2:

According to the Chain Rule, we multiply the two results:

$f'(x) = \left[ \frac{1}{2}(x^2 - 5x)^{-1/2} \right] \cdot (2x - 5)$

To simplify and match the options, we can rewrite the term with the negative exponent as a fraction:

$(x^2 - 5x)^{-1/2} = \frac{1}{(x^2 - 5x)^{1/2}} = \frac{1}{\sqrt{x^2 - 5x}}$

Substituting this back into our derivative expression:

$f'(x) = \frac{1}{2} \cdot \frac{1}{\sqrt{x^2 - 5x}} \cdot (2x - 5)$

Combining this into a single fraction gives:

$f'(x) = \frac{2x - 5}{2\sqrt{x^2 - 5x}}$

This result matches option (C). It is also mathematically identical to option (D).

The correct option is (B) One-one.

Explanation:

This question asks about the property of a function that can be determined using the Horizontal Line Test.

The Horizontal Line Test:

The Horizontal Line Test states that if any horizontal line drawn on the graph of a function intersects the graph at most once, then the function is one-one (or injective).

A one-one function is a function where every distinct input value ($x$) corresponds to a distinct output value ($y$). If a horizontal line (which represents a constant y-value) intersects the graph more than once, it means that multiple x-values are producing the same y-value, so the function is not one-one.

Let's analyze the other options:

(A) Onto: The onto (surjective) property means that the function's range is equal to its codomain. This cannot be determined by the horizontal line test alone. For example, $f(x) = e^x$ passes the horizontal line test (so it's one-one), but its range is $(0, \infty)$, so it is not onto if the codomain is $\mathbb{R}$.

(C) Continuous: Continuity means the graph has no breaks, holes, or jumps. The horizontal line test does not provide information about continuity.

(D) Differentiable: Differentiability means the graph is smooth and has no sharp corners or cusps. The horizontal line test does not provide information about differentiability.

Therefore, passing the horizontal line test is the graphical test for the one-one property.

The correct option is (A) $\{-1, 1\}$.

Explanation:

The range of a function is the set of all possible output values that it can produce.

The function is given by $f(x) = \frac{x}{|x|}$. This is the reciprocal of the function from Question 65, but its behavior is identical. Let's analyze it for different values of $x$.

First, the function is undefined at $x=0$ because the denominator $|x|$ would be zero, leading to division by zero. Thus, 0 is not in the domain.

Now, let's consider the two possible cases for $x \neq 0$:

Case 1: $x$ is positive ($x > 0$)

If $x$ is a positive number, then by definition, $|x| = x$.

Substituting this into the function:

$f(x) = \frac{x}{x} = 1$

For any positive input, the output is always 1.

Case 2: $x$ is negative ($x < 0$)

If $x$ is a negative number, then by definition, $|x| = -x$.

Substituting this into the function:

$f(x) = \frac{x}{-x} = -1$

For any negative input, the output is always -1.

Since the function is not defined for $x=0$, the only possible output values are 1 and -1. Therefore, the range of the function is the discrete set containing just these two values.

In set notation, this is written as $\{-1, 1\}$.

The correct option is (C) $1/2$.

Explanation:

We need to evaluate the limit: $\lim\limits_{x \to 0} \frac{\sqrt{1+x} - 1}{x}$.

If we try to use direct substitution by plugging in $x=0$, we get:

$\frac{\sqrt{1+0} - 1}{0} = \frac{1 - 1}{0} = \frac{0}{0}$

This is an indeterminate form, which means we must manipulate the expression to find the limit.

A common method for limits involving square roots is to multiply the numerator and the denominator by the conjugate of the numerator. The conjugate of $\sqrt{1+x} - 1$ is $\sqrt{1+x} + 1$.

$\lim\limits_{x \to 0} \left( \frac{\sqrt{1+x} - 1}{x} \cdot \frac{\sqrt{1+x} + 1}{\sqrt{1+x} + 1} \right)$

Now, we multiply the numerators. Using the difference of squares formula, $(a-b)(a+b) = a^2 - b^2$:

Numerator: $(\sqrt{1+x})^2 - (1)^2 = (1+x) - 1 = x$

Denominator: $x(\sqrt{1+x} + 1)$

Substitute these back into the limit expression:

$\lim\limits_{x \to 0} \frac{x}{x(\sqrt{1+x} + 1)}$

Since $x$ is approaching 0 but is not equal to 0, we can cancel the $x$ term from the numerator and the denominator:

$\lim\limits_{x \to 0} \frac{1}{\sqrt{1+x} + 1}$

Now we can evaluate the simplified limit by direct substitution:

$\frac{1}{\sqrt{1+0} + 1} = \frac{1}{\sqrt{1} + 1} = \frac{1}{1 + 1} = \frac{1}{2}$

Therefore, the limit is $1/2$.

Alternate Solution (Using L'Hôpital's Rule):

Since we have the indeterminate form $\frac{0}{0}$, we can apply L'Hôpital's Rule, which involves taking the derivative of the numerator and the denominator separately.

Let $f(x) = \sqrt{1+x} - 1 = (1+x)^{1/2} - 1$ and $g(x) = x$.

$f'(x) = \frac{1}{2}(1+x)^{-1/2} \cdot 1 = \frac{1}{2\sqrt{1+x}}$

$g'(x) = 1$

Now, we find the limit of the ratio of the derivatives:

$\lim\limits_{x \to 0} \frac{f'(x)}{g'(x)} = \lim\limits_{x \to 0} \frac{\frac{1}{2\sqrt{1+x}}}{1} = \lim\limits_{x \to 0} \frac{1}{2\sqrt{1+x}}$

By direct substitution:

$\frac{1}{2\sqrt{1+0}} = \frac{1}{2\sqrt{1}} = \frac{1}{2}$

The correct option is (B) 12.

Explanation:

The instantaneous rate of change of a function at a specific point is given by the value of its derivative at that point.

The process involves two steps:

Step 1: Find the derivative function, $f'(x)$.

The given function is $f(x) = x^3$.

We use the power rule for differentiation, which states that $\frac{d}{dx}(x^n) = nx^{n-1}$.

For $f(x) = x^3$, we have $n=3$.

$f'(x) = 3x^{3-1} = 3x^2$

The derivative function, which gives the instantaneous rate of change for any value of $x$, is $f'(x) = 3x^2$.

Step 2: Evaluate the derivative at the specified point, $x=2$.

To find the instantaneous rate of change at $x=2$, we substitute $x=2$ into the derivative function $f'(x)$.

$f'(2) = 3(2)^2$

$f'(2) = 3(4)$

$f'(2) = 12$

Therefore, the instantaneous rate of change of the function $f(x) = x^3$ at the point $x=2$ is 12.

The correct option is (D) $\mathbb{R}$.

Explanation:

The domain of a function is the set of all possible input values (x-values) for which the function is defined and produces a real number as output.

A polynomial function is a function that consists of terms involving only non-negative integer powers of the variable $x$, with each term having a real number coefficient. The general form is $f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$.

Let's consider the operations involved in a polynomial:

• Powers: Raising a real number to a non-negative integer power always results in a real number.
• Multiplication: Multiplying a real number by a constant coefficient ($a_i$) always results in a real number.
• Addition/Subtraction: Adding or subtracting real numbers always results in a real number.

There are no operations in a polynomial that restrict the possible input values. For example, there is no division by a variable (which would cause issues when the denominator is zero) and no square roots of a variable (which would cause issues for negative inputs). You can substitute any real number for $x$ (positive, negative, or zero) into a polynomial and get a valid real number as the output.

Therefore, the domain of any polynomial function is the set of all real numbers, which is denoted by $\mathbb{R}$ or $(-\infty, \infty)$.

The correct option is (C) $f(x) = \frac{x^2 + 1}{x - 3}$.

Explanation:

A rational function is defined as a function that can be expressed as the ratio of two polynomial functions, say $f(x) = \frac{P(x)}{Q(x)}$, where $P(x)$ and $Q(x)$ are polynomials and the denominator polynomial $Q(x)$ is not the zero polynomial.

Let's analyze each of the given options:

(A) $f(x) = \sqrt{x}$: This is a square root function. It can be written as $f(x) = x^{1/2}$. Since the exponent (1/2) is not a non-negative integer, this is not a polynomial function, and therefore, it is not a rational function.

(B) $f(x) = |x|$: This is the absolute value function. It is a piecewise function defined as $f(x) = x$ for $x \ge 0$ and $f(x) = -x$ for $x < 0$. While its pieces are polynomials, the function as a whole cannot be represented by a single ratio of polynomials. It is not a rational function.

(C) $f(x) = \frac{x^2 + 1}{x - 3}$: This function is a ratio of two polynomials:

• The numerator is $P(x) = x^2 + 1$, which is a polynomial.
• The denominator is $Q(x) = x - 3$, which is also a polynomial.

(D) $f(x) = 2^x$: This is an exponential function, where the variable $x$ is in the exponent. This is not a polynomial function, and therefore, not a rational function.

The correct option is (B) $n(x+c)^{n-1} \times 1$. (Note: Option (A) is the simplified form and is also correct).

Explanation:

The function $f(x) = (x+c)^n$ is a composite function, where $c$ and $n$ are constants. We use the Chain Rule to find its derivative.

The Chain Rule states that the derivative is the derivative of the outer function multiplied by the derivative of the inner function.

Let's identify the parts:

• The outer function is the power function, $(\cdot)^n$.
• The inner function is the expression inside the parentheses, $x+c$.

Now, we differentiate step-by-step:

1. Differentiate the outer function:

Using the power rule, the derivative of $(\cdot)^n$ is $n(\cdot)^{n-1}$. We keep the inner function, $(x+c)$, unchanged inside:

$n(x+c)^{n-1}$

2. Differentiate the inner function:

The derivative of the inner function, $x+c$, with respect to $x$ is:

$\frac{d}{dx}(x+c) = 1 + 0 = 1$ (since $c$ is a constant, its derivative is 0).

3. Multiply the results from Step 1 and Step 2:

According to the Chain Rule, we multiply the two results:

$f'(x) = \underbrace{n(x+c)^{n-1}}_{\text{Derivative of outer}} \cdot \underbrace{(1)}_{\text{Derivative of inner}}$

So, the derivative is $n(x+c)^{n-1} \times 1$.

This matches option (B). Although multiplying by 1 does not change the value, option (B) explicitly shows both parts of the chain rule calculation. Option (A) is the simplified result.

The correct option is (C) continuity and differentiation.

Explanation:

The concept of a limit is the cornerstone of calculus and is used to formally define several other key concepts.

Let's analyze the options:

(A) functions and relations: These are more fundamental concepts in set theory and algebra. The idea of a function or a relation does not require the concept of a limit.

(B) domain and range: These are properties of a function, describing its possible inputs and outputs. They are defined based on the function's rule and context, not directly on the concept of a limit.

(C) continuity and differentiation: This is the correct answer. Both of these core calculus concepts are defined using limits:

• Continuity: A function $f(x)$ is continuous at a point $x=a$ if and only if $\lim\limits_{x \to a} f(x) = f(a)$. This definition explicitly uses a limit.
• Differentiation: The derivative of a function $f(x)$ at a point $x=a$ is defined as the limit of the difference quotient: $f'(a) = \lim\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$. This definition is also explicitly based on a limit.

(D) sequences and series: While the convergence of sequences and series is defined using limits (e.g., the sum of an infinite series is the limit of its partial sums), the concepts of continuity and differentiation are the most direct and foundational applications of the limit of a function in introductory calculus.

Solution:

The experiment described consists of two stages. First, a coin is tossed. The outcome of the toss determines the second stage of the experiment.

Let H be the event of getting a Head and T be the event of getting a Tail on the coin toss.

Case 1: The coin shows a Head (H)

When a head occurs, we draw a ball from a bag. The bag contains 3 blue balls and 4 white balls. To list all possible outcomes, we need to consider each ball as a distinct entity.

Let the 3 blue balls be denoted by $B_1, B_2, B_3$.

Let the 4 white balls be denoted by $W_1, W_2, W_3, W_4$.

The set of outcomes when the coin shows a head is:

$\{H B_1, H B_2, H B_3, H W_1, H W_2, H W_3, H W_4\}$

Case 2: The coin shows a Tail (T)

When a tail occurs, we throw a standard six-sided die. The possible outcomes from throwing a die are the numbers from 1 to 6.

The set of outcomes when the coin shows a tail is:

$\{T1, T2, T3, T4, T5, T6\}$

Final Sample Space

The total sample space (S) of the experiment is the collection of all possible outcomes from both cases.

By combining the outcomes from Case 1 and Case 2, we get the complete sample space:

$S = \{H B_1, H B_2, H B_3, H W_1, H W_2, H W_3, H W_4, T1, T2, T3, T4, T5, T6\}$

Alternate Solution:

If we consider the balls of the same color to be indistinguishable (i.e., we are only concerned with the color of the ball drawn, not the specific ball), the sample space can be described more concisely.

Let B represent the event of drawing a blue ball and W represent the event of drawing a white ball.

Case 1: Coin shows Head (H)

The outcomes are simply drawing a blue ball or a white ball. The associated sample points are $\{HB, HW\}$.

Case 2: Coin shows Tail (T)

This part remains unchanged. The outcomes are $\{T1, T2, T3, T4, T5, T6\}$.

Final Sample Space (Alternate)

The combined sample space S under this assumption is:

$S = \{HB, HW, T1, T2, T3, T4, T5, T6\}$

Note: The first solution, where each ball is treated as a distinct item, is generally preferred for probability calculations as it creates a sample space of equally likely elementary outcomes.

Given:

The function is given by $f(x) = 2x + 3$.

To Find:

The domain and range of the function $f(x)$.

Solution:

The given function is a linear polynomial function, $f(x) = 2x + 3$.

Domain of f(x)

The domain of a function is the set of all possible input values (x-values) for which the function is defined. For the function $f(x) = 2x + 3$, we can see that the expression $2x + 3$ is defined for all real numbers. There are no values of $x$ that would lead to an undefined result (like division by zero or the square root of a negative number).

Therefore, the domain of $f(x)$ is the set of all real numbers.

Domain = $\mathbb{R}$ or $(-\infty, \infty)$.

Range of f(x)

The range of a function is the set of all possible output values (y-values or f(x)-values). To find the range, let's set $y = f(x)$.

$y = 2x + 3$

Now, let's solve this equation for $x$ in terms of $y$:

$y - 3 = 2x$

$x = \frac{y - 3}{2}$

This equation shows that for any real value of $y$, we can find a corresponding real value of $x$. This means that $y$ can take any real value.

Therefore, the range of $f(x)$ is the set of all real numbers.

Range = $\mathbb{R}$ or $(-\infty, \infty)$.

Given:

The function is given by $f(x) = \frac{1}{x-2}$.

To Find:

The domain of the function $f(x)$.

Solution:

The domain of a function is the set of all possible input values (x-values) for which the function is defined.

The given function $f(x) = \frac{1}{x-2}$ is a rational function. A rational function is defined everywhere except for the values of $x$ that make the denominator equal to zero, as division by zero is undefined.

To find the value(s) of $x$ for which the function is not defined, we set the denominator equal to zero:

$x - 2 = 0$

Solving this simple equation for $x$, we get:

$x = 2$

This means the function $f(x)$ is not defined at $x=2$. For all other real numbers, the denominator is non-zero, and the function yields a real number.

Therefore, the domain of the function $f(x)$ is the set of all real numbers except for 2.

We can express the domain in two common ways:

1. Set-builder notation: Domain = $\{x \in \mathbb{R} \mid x \neq 2\}$

2. Interval notation: Domain = $(-\infty, 2) \cup (2, \infty)$

Given:

The function is given by $f(x) = \sqrt{x-4}$.

To Find:

The domain of the function $f(x)$.

Solution:

The domain of a function is the set of all possible input values (x-values) for which the function is defined in the set of real numbers.

The given function is $f(x) = \sqrt{x-4}$. This function involves a square root. For the square root of an expression to be a real number, the expression inside the square root (called the radicand) must be non-negative (i.e., greater than or equal to zero).

Therefore, for $f(x)$ to be defined, we must have:

Now, we solve this inequality for $x$ by adding 4 to both sides:

This result tells us that the function $f(x)$ is defined for all real numbers $x$ that are greater than or equal to 4.

Therefore, the domain of the function is the set of all real numbers from 4 to infinity, inclusive of 4.

We can express the domain in two ways:

1. Set-builder notation: Domain = $\{x \in \mathbb{R} \mid x \ge 4\}$

2. Interval notation: Domain = $[4, \infty)$

Given:

The function is $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^2$.

The domain is the set of all real numbers, $\mathbb{R}$.

The codomain is the set of all real numbers, $\mathbb{R}$.

To Classify:

Determine if the function $f(x)$ is one-one (injective), onto (surjective), or neither.

Justification:

We will check the properties of being one-one and onto separately.

1. Check for One-One (Injective)

A function is one-one if different elements in the domain have different images in the codomain. That is, if $f(x_1) = f(x_2)$, then it must be that $x_1 = x_2$.

Let's test this with a counterexample. Consider two different elements from the domain $\mathbb{R}$, for example, $x_1 = -2$ and $x_2 = 2$.

Calculate their images under $f(x)$:

$f(x_1) = f(-2) = (-2)^2 = 4$

$f(x_2) = f(2) = (2)^2 = 4$

Here, we have $f(-2) = f(2)$, but $-2 \neq 2$.

Since two different elements in the domain ($-2$ and $2$) map to the same element in the codomain (4), the function is not one-one.

2. Check for Onto (Surjective)

A function is onto if every element in the codomain has at least one pre-image in the domain. This means the range of the function must be equal to its codomain.

The codomain of $f(x) = x^2$ is given as $\mathbb{R}$.

Let's find the range of the function. The range is the set of all possible output values. For any real number $x$, its square, $x^2$, is always non-negative.

$x^2 \ge 0$

So, the range of $f(x) = x^2$ is the set of all non-negative real numbers, which is $[0, \infty)$.

Now we compare the range with the codomain:

Range = $[0, \infty)$

Codomain = $\mathbb{R} = (-\infty, \infty)$

Since the Range $\neq$ Codomain, the function is not onto. For example, there is no real number $x$ in the domain such that $f(x) = x^2 = -1$, because $-1$ is in the codomain but not in the range.

Conclusion:

Since the function $f(x) = x^2$ is neither one-one nor onto, we classify it as neither one-one nor onto.

Definition of a Constant Function

A constant function is a function whose output value is the same for every input value. In other words, if $f$ is a function from a set A to a set B, then $f$ is a constant function if there is a fixed element $c$ in B such that $f(x) = c$ for all $x$ in A.

Example of a Constant Function

Let's consider the function $f: \mathbb{R} \to \mathbb{R}$ defined by:

$f(x) = 3$

This is a constant function because no matter what real number we choose for the input $x$, the output is always 3.

• If $x = -5$, then $f(-5) = 3$.
• If $x = 0$, then $f(0) = 3$.
• If $x = 10$, then $f(10) = 3$.

The domain of this function is the set of all real numbers, $\mathbb{R}$.

The range of this function is the single value {3}.

Graph of the Constant Function $f(x) = 3$

The graph of a function is the set of all ordered pairs $(x, f(x))$. For our example, the ordered pairs are of the form $(x, 3)$. Some points on the graph are:

$(-2, 3), (-1, 3), (0, 3), (1, 3), (2, 3)$, etc.

When we plot these points on a Cartesian coordinate system, they form a horizontal line.

Description of the graph:

• The graph is a straight line that is parallel to the x-axis.
• The line is located 3 units above the x-axis.
• The y-intercept of the line is at the point (0, 3). For every x-value, the y-value is 3.

Here is a textual representation of the sketch:

^ y-axis
|
4 +
|
3 +---------------------------------- <-- Graph of f(x) = 3
|
2 +
|
1 +
+---+---+---+---+---+---+--> x-axis
-3 -2 -1 0 1 2 3

Given:

We need to evaluate the limit:

$\lim\limits_{x \to 3} (2x^2 - 5x + 1)$

To Find:

The value of the given limit.

Solution:

The function given is $f(x) = 2x^2 - 5x + 1$, which is a polynomial function.

For any polynomial function $p(x)$, the limit as $x$ approaches a real number $a$ can be found by direct substitution, i.e., $\lim\limits_{x \to a} p(x) = p(a)$.

This is because polynomial functions are continuous everywhere.

So, to find the limit, we can substitute $x=3$ directly into the expression:

$\lim\limits_{x \to 3} (2x^2 - 5x + 1) = 2(3)^2 - 5(3) + 1$

Now, we simplify the expression:

$= 2(9) - 15 + 1$

$= 18 - 15 + 1$

$= 3 + 1$

$= 4$

Therefore, the value of the limit is 4.

Final Answer: $\lim\limits_{x \to 3} (2x^2 - 5x + 1) = 4$

Alternate Solution (Using Limit Properties):

We can also solve this by applying the properties of limits (sum, difference, and constant multiple rules).

The limit of a sum/difference is the sum/difference of the limits:

$\lim\limits_{x \to 3} (2x^2 - 5x + 1) = \lim\limits_{x \to 3} (2x^2) - \lim\limits_{x \to 3} (5x) + \lim\limits_{x \to 3} (1)$

Now, apply the constant multiple rule $\lim\limits_{x \to a} [c \cdot f(x)] = c \cdot \lim\limits_{x \to a} f(x)$:

$= 2 \lim\limits_{x \to 3} (x^2) - 5 \lim\limits_{x \to 3} (x) + \lim\limits_{x \to 3} (1)$

Evaluate each limit:

$= 2 (3^2) - 5(3) + 1$

Simplify the expression:

$= 2(9) - 15 + 1$

$= 18 - 15 + 1$

$= 4$

Both methods yield the same result.

Given:

We need to evaluate the limit:

$\lim\limits_{x \to 2} \frac{x^2 - 4}{x - 2}$

To Find:

The value of the given limit.

Solution:

First, we try to evaluate the limit by direct substitution. If we substitute $x = 2$ into the expression, we get:

Numerator: $(2)^2 - 4 = 4 - 4 = 0$

Denominator: $2 - 2 = 0$

This results in the indeterminate form $\frac{0}{0}$. This means we cannot find the limit by direct substitution and must simplify the expression first.

We can simplify the expression by factoring the numerator. The numerator, $x^2 - 4$, is a difference of squares, which can be factored using the identity $a^2 - b^2 = (a - b)(a + b)$.

$x^2 - 4 = (x)^2 - (2)^2 = (x - 2)(x + 2)$

Now, we can substitute this back into the limit expression:

$\lim\limits_{x \to 2} \frac{(x - 2)(x + 2)}{x - 2}$

Since the limit is taken as $x$ approaches 2, $x$ is not equal to 2. Therefore, $x - 2 \neq 0$, and we can cancel the $(x - 2)$ term from the numerator and the denominator.

$\lim\limits_{x \to 2} \frac{\cancel{(x - 2)}(x + 2)}{\cancel{x - 2}} = \lim\limits_{x \to 2} (x + 2)$

Now, we can evaluate the simplified limit by direct substitution:

$\lim\limits_{x \to 2} (x + 2) = 2 + 2 = 4$

Therefore, the value of the limit is 4.

Final Answer: $\lim\limits_{x \to 2} \frac{x^2 - 4}{x - 2} = 4$

Alternate Solution (Using L'Hôpital's Rule)

Since the limit results in the indeterminate form $\frac{0}{0}$ upon direct substitution, we can apply L'Hôpital's Rule.

L'Hôpital's Rule states that if $\lim\limits_{x \to a} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim\limits_{x \to a} \frac{f(x)}{g(x)} = \lim\limits_{x \to a} \frac{f'(x)}{g'(x)}$, provided the limit on the right exists.

Let $f(x) = x^2 - 4$ and $g(x) = x - 2$.

Find the derivatives of $f(x)$ and $g(x)$:

$f'(x) = \frac{d}{dx}(x^2 - 4) = 2x$

$g'(x) = \frac{d}{dx}(x - 2) = 1$

Now, apply the rule:

$\lim\limits_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim\limits_{x \to 2} \frac{f'(x)}{g'(x)} = \lim\limits_{x \to 2} \frac{2x}{1}$

Now, substitute $x=2$ into the new expression:

$\lim\limits_{x \to 2} \frac{2x}{1} = \frac{2(2)}{1} = 4$

Both methods yield the same result.

Given:

A trapezium ABCD in which the side AB is parallel to the side DC (AB || DC).

The diagonals AC and BD intersect each other at point O.

To Prove:

$\frac{AO}{BO} = \frac{CO}{DO}$

Proof:

We will use the concept of similar triangles to prove the given statement.

Consider the triangles $\triangle AOB$ and $\triangle COD$.

Thus, by the Angle-Angle-Angle (AAA) similarity criterion, we have:

$\triangle AOB \sim \triangle COD$

Since the two triangles are similar, the ratio of their corresponding sides must be equal.

$\frac{AO}{CO} = \frac{BO}{DO} = \frac{AB}{CD}$

Now, let's consider the first two parts of the proportion:

$\frac{AO}{CO} = \frac{BO}{DO}$

To get the desired form, we can rearrange the terms (using alternendo property of proportions):

$\frac{AO}{BO} = \frac{CO}{DO}$

Hence, Proved.

Alternate Solution (Using Basic Proportionality Theorem)

Construction:

Draw a line segment EF through the point O such that EF || AB. Since AB || DC, we have EF || DC.

Proof:

First, consider the triangle $\triangle ADC$.

In $\triangle ADC$, we have EO || DC.

By the Basic Proportionality Theorem (BPT), the line EO divides the sides AD and AC in the same ratio.

Now, consider the triangle $\triangle DAB$.

In $\triangle DAB$, we have EO || AB.

By the Basic Proportionality Theorem (BPT), the line EO divides the sides AD and BD in the same ratio.

From equation (i) and equation (ii), we can equate the expressions for $\frac{AE}{ED}$:

$\frac{AO}{CO} = \frac{BO}{DO}$

By rearranging the terms, we get:

$\frac{AO}{BO} = \frac{CO}{DO}$

Hence, Proved.

To determine if the function $f(x) = x^3$ is continuous at $x=1$, we need to check if it satisfies the three conditions for continuity at a point.

A function $f(x)$ is said to be continuous at a point $x=c$ if:

1. $f(c)$ is defined.

2. The limit of the function as $x$ approaches $c$ exists. This means the Left-Hand Limit (LHL) and the Right-Hand Limit (RHL) are equal.

$\lim\limits_{x \to c^-} f(x) = \lim\limits_{x \to c^+} f(x)$

3. The value of the function at the point is equal to the limit of the function at that point.

$\lim\limits_{x \to c} f(x) = f(c)$

Given:

The function is $f(x) = x^3$.

The point is $x=1$.

Solution:

We will now check the three conditions for continuity at $x=1$.

Step 1: Find the value of the function at $x=1$.

Substitute $x=1$ into the function:

$f(1) = (1)^3 = 1$

The function is defined at $x=1$.

Step 2: Find the Left-Hand Limit (LHL) at $x=1$.

The LHL is given by $\lim\limits_{x \to 1^-} f(x) = \lim\limits_{h \to 0} f(1-h)$.

$\lim\limits_{h \to 0} (1-h)^3$

Applying the limit by substituting $h=0$:

$(1-0)^3 = 1^3 = 1$

So, LHL = 1.

Step 3: Find the Right-Hand Limit (RHL) at $x=1$.

The RHL is given by $\lim\limits_{x \to 1^+} f(x) = \lim\limits_{h \to 0} f(1+h)$.

$\lim\limits_{h \to 0} (1+h)^3$

Applying the limit by substituting $h=0$:

$(1+0)^3 = 1^3 = 1$

So, RHL = 1.

Step 4: Compare the values.

From our calculations:

LHL = 1

RHL = 1

$f(1) = 1$

Since $\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^+} f(x) = f(1)$, all three conditions for continuity are satisfied.

Conclusion:

Yes, the function $f(x) = x^3$ is continuous at $x=1$ because the limit of the function as $x$ approaches 1 exists and is equal to the value of the function at $x=1$.

Definition of Instantaneous Rate of Change

The instantaneous rate of change of a function $f(x)$ at a specific point $x=a$ describes how fast the function's output value is changing at that precise instant.

It is derived from the concept of the average rate of change. The average rate of change over a small interval, from $x=a$ to $x=a+h$, is the slope of the secant line connecting the points $(a, f(a))$ and $(a+h, f(a+h))$.

Average Rate of Change = $\frac{\text{Change in } f(x)}{\text{Change in } x} = \frac{f(a+h) - f(a)}{(a+h) - a} = \frac{f(a+h) - f(a)}{h}$

To find the rate of change at the single instant $x=a$, we imagine the interval $h$ becoming infinitesimally small. This is achieved by taking the limit of the average rate of change as $h$ approaches 0.

Therefore, the formal definition of the instantaneous rate of change of $f(x)$ at $x=a$ is given by the limit:

Instantaneous Rate of Change at $a = \lim\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$

Relationship to the Derivative

The instantaneous rate of change of a function $f(x)$ at a point $x=a$ is identical to the definition of the derivative of the function $f(x)$ at that same point, denoted as $f'(a)$.

The definition of the derivative of $f(x)$ at $x=a$ (using the limit definition) is:

$f'(a) = \lim\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$

As you can see, the formulas are exactly the same. Thus, we can conclude that:

Instantaneous Rate of Change at $x=a$ = The Derivative at $x=a$ ($f'(a)$)

In essence, "instantaneous rate of change" and "derivative" are two different terms for the same mathematical concept:

• The term "instantaneous rate of change" emphasizes the physical or practical application, such as finding the instantaneous velocity of an object from its position function.
• The term "derivative" is the formal mathematical term, which also has a powerful geometric interpretation as the slope of the tangent line to the graph of the function at that point.

To find the derivative of a function using the definition of the derivative, also known as the first principles method, we use the following limit formula:

Given:

The function is $f(x) = x^4$.

Solution:

Step 1: Determine the expression for $f(x+h)$.

Substitute $(x+h)$ into the function $f(x)$:

$f(x+h) = (x+h)^4$

Step 2: Expand the expression for $f(x+h)$.

We use the Binomial Theorem to expand $(x+h)^4$:

$(a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$

Substituting $a=x$ and $b=h$:

$(x+h)^4 = x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4$

Step 3: Substitute $f(x+h)$ and $f(x)$ into the derivative formula.

$f'(x) = \lim\limits_{h \to 0} \frac{(x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4) - x^4}{h}$

Step 4: Simplify the numerator.

The $x^4$ terms cancel each other out:

$f'(x) = \lim\limits_{h \to 0} \frac{4x^3h + 6x^2h^2 + 4xh^3 + h^4}{h}$

Step 5: Factor out $h$ from the numerator.

Each term in the numerator has a factor of $h$:

$f'(x) = \lim\limits_{h \to 0} \frac{h(4x^3 + 6x^2h + 4xh^2 + h^3)}{h}$

Step 6: Cancel the $h$ term.

We can cancel $h$ from the numerator and the denominator, as $h \to 0$ but is not equal to 0.

$f'(x) = \lim\limits_{h \to 0} (4x^3 + 6x^2h + 4xh^2 + h^3)$

Step 7: Evaluate the limit by substituting $h=0$.

Now we substitute $h=0$ into the simplified expression:

$f'(x) = 4x^3 + 6x^2(0) + 4x(0)^2 + (0)^3$

$f'(x) = 4x^3 + 0 + 0 + 0$

$f'(x) = 4x^3$

Thus, the derivative of $f(x) = x^4$ is $f'(x) = 4x^3$.

To find the derivative of the function $f(x) = 3x^2 - 5x + 7$, we will use the standard rules of differentiation, which allow us to find the derivative without using the limit definition (first principles).

The key rules we will use are:

• The Power Rule: For any real number $n$, the derivative of $x^n$ is given by $\frac{d}{dx}(x^n) = nx^{n-1}$.
• The Constant Multiple Rule: The derivative of a constant multiplied by a function is the constant times the derivative of the function: $\frac{d}{dx}(c \cdot g(x)) = c \cdot \frac{d}{dx}(g(x))$.
• The Sum/Difference Rule: The derivative of a sum or difference of functions is the sum or difference of their derivatives: $\frac{d}{dx}(g(x) \pm h(x)) = \frac{d}{dx}(g(x)) \pm \frac{d}{dx}(h(x))$.
• The Constant Rule: The derivative of any constant number is 0: $\frac{d}{dx}(c) = 0$.

Given:

The function is $f(x) = 3x^2 - 5x + 7$.

Solution:

We need to find the derivative of $f(x)$, which is denoted as $f'(x)$ or $\frac{df}{dx}$.

$f'(x) = \frac{d}{dx}(3x^2 - 5x + 7)$

Step 1: Apply the Sum/Difference Rule.

We differentiate the function term by term:

$f'(x) = \frac{d}{dx}(3x^2) - \frac{d}{dx}(5x) + \frac{d}{dx}(7)$

Step 2: Differentiate each term individually.

For the first term, $\frac{d}{dx}(3x^2)$:

Using the Constant Multiple Rule and the Power Rule (where $n=2$):

$\frac{d}{dx}(3x^2) = 3 \cdot \frac{d}{dx}(x^2) = 3 \cdot (2x^{2-1}) = 3 \cdot (2x) = 6x$

For the second term, $\frac{d}{dx}(5x)$:

We can write $5x$ as $5x^1$. Using the Constant Multiple Rule and the Power Rule (where $n=1$):

$\frac{d}{dx}(5x^1) = 5 \cdot \frac{d}{dx}(x^1) = 5 \cdot (1x^{1-1}) = 5 \cdot (1x^0) = 5 \cdot 1 = 5$

For the third term, $\frac{d}{dx}(7)$:

Using the Constant Rule, the derivative of a constant is 0:

$\frac{d}{dx}(7) = 0$

Step 3: Combine the results.

Substitute the derivatives of each term back into the expression from Step 1:

$f'(x) = 6x - 5 + 0$

$f'(x) = 6x - 5$

Therefore, the derivative of the function $f(x) = 3x^2 - 5x + 7$ is $f'(x) = 6x - 5$.

To find the derivative of the composite function $y = (x^2 + 1)^3$, we use the Chain Rule.

The Chain Rule states that if a function $y$ is a composition of two other functions, say $y = f(u)$ and $u = g(x)$, then the derivative of $y$ with respect to $x$ is given by:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

In simpler terms, we differentiate the "outer function" and multiply it by the derivative of the "inner function".

Given:

The function is $y = (x^2 + 1)^3$.

Solution:

Step 1: Identify the inner and outer functions.

Let the "inner function" be $u$.

$u = x^2 + 1$

Now, the function $y$ can be written in terms of $u$ as the "outer function":

$y = u^3$

Step 2: Find the derivative of the outer function ($\frac{dy}{du}$).

We differentiate $y = u^3$ with respect to $u$ using the Power Rule:

$\frac{dy}{du} = \frac{d}{du}(u^3) = 3u^{3-1} = 3u^2$

Step 3: Find the derivative of the inner function ($\frac{du}{dx}$).

We differentiate $u = x^2 + 1$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(x^2 + 1) = \frac{d}{dx}(x^2) + \frac{d}{dx}(1)$

$\frac{du}{dx} = 2x + 0 = 2x$

Step 4: Apply the Chain Rule formula.

Multiply the results from Step 2 and Step 3:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

$\frac{dy}{dx} = (3u^2) \cdot (2x)$

Step 5: Substitute the expression for $u$ back into the derivative.

Replace $u$ with its original expression, $x^2 + 1$:

$\frac{dy}{dx} = 3(x^2 + 1)^2 \cdot (2x)$

Step 6: Simplify the final expression.

Multiply the constants and rearrange the terms for the final form:

$\frac{dy}{dx} = 6x(x^2 + 1)^2$

Therefore, the derivative of $y = (x^2 + 1)^3$ is $\frac{dy}{dx} = 6x(x^2 + 1)^2$.

The range of a function is the set of all possible output values (y-values) that the function can produce from its given domain.

Given:

The function is $f(x) = |x|$.

To Find:

The range of the function $f(x)$.

Solution:

The absolute value function, $f(x) = |x|$, gives the magnitude or distance of a number from zero on the number line, regardless of its sign. The formal definition is:

$f(x) = |x| = \begin{cases} x & , & \text{if } x \geq 0 \\ -x & , & \text{if } x < 0 \end{cases}$

We can analyze the output of the function for all possible types of input values of $x$ (which can be any real number):

• When $x$ is a positive number (e.g., $x=5$): The output is $f(5) = |5| = 5$. The output is positive.
• When $x$ is zero (i.e., $x=0$): The output is $f(0) = |0| = 0$. The output is zero.
• When $x$ is a negative number (e.g., $x=-5$): The output is $f(-5) = |-5| = -(-5) = 5$. The output is positive.

From this analysis, we can conclude that for any real number input $x$, the output $f(x)$ will always be greater than or equal to zero. The function can never produce a negative value because the definition of absolute value ensures a non-negative result.

Thus, the set of all possible output values includes 0 and all positive real numbers.

Conclusion:

The range of the function $f(x) = |x|$ is the set of all non-negative real numbers.

In interval notation, this is written as $[0, \infty)$.

A polynomial function is a function that can be expressed in the form:

$f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$

where $n$ is a non-negative integer (the degree of the polynomial) and $a_n, a_{n-1}, \dots, a_0$ are real number coefficients.

Example of a Polynomial Function

A good example of a polynomial function is a quadratic function. Let's consider the following:

$f(x) = x^2 - 4x + 5$

This is a polynomial of degree 2, and its graph is an upward-opening parabola.

Domain of the Function

The domain of a function is the set of all possible input values (x-values) for which the function is defined.

For any polynomial function, including $f(x) = x^2 - 4x + 5$, there are no restrictions on the input value $x$. We can substitute any real number for $x$ and get a valid output. There are no divisions by zero or square roots of negative numbers to worry about.

Therefore, the domain is all real numbers.

In interval notation, the domain is $(-\infty, \infty)$.

Range of the Function

The range of a function is the set of all possible output values (y-values).

For the quadratic function $f(x) = x^2 - 4x + 5$, the graph is a parabola that opens upwards (because the coefficient of the $x^2$ term, which is 1, is positive). This means the function has a minimum value at its vertex.

To find the vertex of a parabola given by $ax^2 + bx + c$, we use the formula for the x-coordinate: $x = -\frac{b}{2a}$.

In our function, $a=1$, $b=-4$, and $c=5$.

x-coordinate of the vertex = $-\frac{(-4)}{2(1)} = \frac{4}{2} = 2$

Now, we find the y-coordinate of the vertex (the minimum value of the function) by substituting this x-value back into the function:

$f(2) = (2)^2 - 4(2) + 5$

$f(2) = 4 - 8 + 5$

$f(2) = 1$

The vertex is at the point $(2, 1)$. Since this is the minimum point of the function, the function's output can be 1 or any value greater than 1.

Therefore, the range is all real numbers greater than or equal to 1.

In interval notation, the range is $[1, \infty)$.

To evaluate the limit $\lim\limits_{x \to 0} \frac{\sqrt{x+1} - 1}{x}$, we first attempt to use direct substitution.

Substituting $x=0$ into the expression gives:

$\frac{\sqrt{0+1} - 1}{0} = \frac{\sqrt{1} - 1}{0} = \frac{1-1}{0} = \frac{0}{0}$

This is an indeterminate form, which means we must manipulate the expression algebraically before we can evaluate the limit.

Solution using Conjugate Multiplication

A common method for limits involving square roots is to multiply the numerator and the denominator by the conjugate of the expression containing the square root.

The conjugate of the numerator, $\sqrt{x+1} - 1$, is $\sqrt{x+1} + 1$.

Step 1: Multiply the expression by the conjugate.

$\lim\limits_{x \to 0} \frac{\sqrt{x+1} - 1}{x} \cdot \frac{\sqrt{x+1} + 1}{\sqrt{x+1} + 1}$

Step 2: Simplify the numerator.

We use the difference of squares formula, $(a-b)(a+b) = a^2 - b^2$, where $a = \sqrt{x+1}$ and $b = 1$.

$(\sqrt{x+1} - 1)(\sqrt{x+1} + 1) = (\sqrt{x+1})^2 - (1)^2 = (x+1) - 1 = x$

Substituting this back into the limit expression:

$\lim\limits_{x \to 0} \frac{x}{x(\sqrt{x+1} + 1)}$

Step 3: Cancel the common factor.

Since $x$ is approaching 0 but is not equal to 0, we can cancel the $x$ term from the numerator and the denominator.

$\lim\limits_{x \to 0} \frac{\cancel{x}}{\cancel{x}(\sqrt{x+1} + 1)} = \lim\limits_{x \to 0} \frac{1}{\sqrt{x+1} + 1}$

Step 4: Evaluate the limit by direct substitution.

Now we can substitute $x=0$ into the simplified expression:

$\frac{1}{\sqrt{0+1} + 1} = \frac{1}{\sqrt{1} + 1} = \frac{1}{1+1} = \frac{1}{2}$

Alternate Solution using L'Hôpital's Rule

Since the initial limit is of the indeterminate form $\frac{0}{0}$, we can apply L'Hôpital's Rule. This rule states that we can take the derivative of the numerator and the denominator separately and then evaluate the limit.

Let $f(x) = \sqrt{x+1} - 1$ and $g(x) = x$.

Step 1: Find the derivatives of the numerator and denominator.

$f'(x) = \frac{d}{dx}(\sqrt{x+1} - 1) = \frac{d}{dx}((x+1)^{1/2}) - \frac{d}{dx}(1) = \frac{1}{2}(x+1)^{-1/2} - 0 = \frac{1}{2\sqrt{x+1}}$

$g'(x) = \frac{d}{dx}(x) = 1$

Step 2: Apply L'Hôpital's Rule.

$\lim\limits_{x \to 0} \frac{\sqrt{x+1} - 1}{x} = \lim\limits_{x \to 0} \frac{f'(x)}{g'(x)} = \lim\limits_{x \to 0} \frac{\frac{1}{2\sqrt{x+1}}}{1}$

Step 3: Evaluate the new limit.

$\lim\limits_{x \to 0} \frac{1}{2\sqrt{x+1}} = \frac{1}{2\sqrt{0+1}} = \frac{1}{2\sqrt{1}} = \frac{1}{2}$

Both methods yield the same result.

The value of the limit is $\frac{1}{2}$.

To check if the given piecewise function is continuous at $x=2$, we must verify if it satisfies the three conditions for continuity at a point $x=c$.

A function $f(x)$ is continuous at $x=c$ if:

1. $f(c)$ is defined.
2. The limit $\lim\limits_{x \to c} f(x)$ exists. This requires the Left-Hand Limit (LHL) to be equal to the Right-Hand Limit (RHL).
3. The limit equals the function's value: $\lim\limits_{x \to c} f(x) = f(c)$.

Given:

The function is defined as:

$f(x) = \begin{cases} x+1 & , & \text{if } x \leq 2 \\ 5-x & , & \text{if } x > 2 \end{cases}$

The point to check for continuity is $x=2$.

Solution:

We will check the three conditions for continuity at $x=2$.

Step 1: Find the value of the function at $x=2$.

According to the function's definition, for $x \leq 2$, we use the expression $f(x) = x+1$. Since we are evaluating at exactly $x=2$, this is the part of the function we must use.

$f(2) = 2 + 1 = 3$

The function is defined at $x=2$, and its value is 3.

Step 2: Find the Left-Hand Limit (LHL).

The LHL is the limit as $x$ approaches 2 from values less than 2 ($x \to 2^-$). For this, we use the expression $f(x) = x+1$.

$\text{LHL} = \lim\limits_{x \to 2^-} f(x) = \lim\limits_{x \to 2^-} (x+1)$

Substituting $x=2$ into the expression:

$\text{LHL} = 2 + 1 = 3$

Step 3: Find the Right-Hand Limit (RHL).

The RHL is the limit as $x$ approaches 2 from values greater than 2 ($x \to 2^+$). For this, we use the expression $f(x) = 5-x$.

$\text{RHL} = \lim\limits_{x \to 2^+} f(x) = \lim\limits_{x \to 2^+} (5-x)$

Substituting $x=2$ into the expression:

$\text{RHL} = 5 - 2 = 3$

Step 4: Compare the results.

We have found:

• $f(2) = 3$
• LHL = 3
• RHL = 3

Since $\text{LHL} = \text{RHL} = 3$, the limit exists and is equal to 3.

$\lim\limits_{x \to 2} f(x) = 3$

Furthermore, this limit is equal to the value of the function at the point:

$\lim\limits_{x \to 2} f(x) = f(2)$

Since all three conditions for continuity are satisfied, the function is continuous at $x=2$.

Conclusion:

Yes, the function $f(x)$ is continuous at $x=2$.

To find the derivative of the composite function $y = (2x+3)^5$, we will use the Chain Rule. The Chain Rule is used to differentiate a "function of a function".

The rule states that if $y$ is a function of $u$, and $u$ is a function of $x$, then the derivative of $y$ with respect to $x$ is given by the formula:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

Given:

The function is $y = (2x+3)^5$.

To Find:

The derivative $\frac{dy}{dx}$.

Solution:

We can solve this by breaking the function down into an "inner" part and an "outer" part.

Step 1: Identify the inner function ($u$) and the outer function ($y(u)$).

Let the inner function be $u = 2x+3$.

Then, the outer function becomes $y = u^5$.

Step 2: Differentiate the outer function with respect to $u$ ($\frac{dy}{du}$).

Using the Power Rule on $y=u^5$:

$\frac{dy}{du} = 5u^{5-1} = 5u^4$

Step 3: Differentiate the inner function with respect to $x$ ($\frac{du}{dx}$).

Differentiating $u=2x+3$:

$\frac{du}{dx} = \frac{d}{dx}(2x+3) = \frac{d}{dx}(2x) + \frac{d}{dx}(3) = 2 + 0 = 2$

Step 4: Apply the Chain Rule.

Multiply the results from Step 2 and Step 3 according to the Chain Rule formula:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

$\frac{dy}{dx} = (5u^4) \cdot (2)$

Step 5: Substitute the expression for $u$ back into the equation.

Replace $u$ with its original expression, $2x+3$:

$\frac{dy}{dx} = 5(2x+3)^4 \cdot 2$

Step 6: Simplify the final expression.

$\frac{dy}{dx} = 10(2x+3)^4$

Therefore, the derivative of $y = (2x+3)^5$ is $\frac{dy}{dx} = 10(2x+3)^4$.

This question has two parts: calculating the average rate of change and explaining how it differs from the instantaneous rate of change.

Part 1: Average Rate of Change Calculation

The average rate of change of a function $f(x)$ over an interval from $x=a$ to $x=b$ is the slope of the secant line connecting the two points $(a, f(a))$ and $(b, f(b))$. It is calculated using the formula:

Average Rate of Change = $\frac{f(b) - f(a)}{b-a}$

Given:

The function is $f(x) = x^2$.

The interval is from $x=1$ to $x=3$. So, $a=1$ and $b=3$.

To Find:

The average rate of change of $f(x)$ on the interval $[1, 3]$.

Solution:

Step 1: Evaluate the function at the endpoints of the interval.

Find the value of $f(a) = f(1)$:

$f(1) = (1)^2 = 1$

Find the value of $f(b) = f(3)$:

$f(3) = (3)^2 = 9$

Step 2: Substitute these values into the average rate of change formula.

Average Rate of Change = $\frac{f(3) - f(1)}{3 - 1}$

Average Rate of Change = $\frac{9 - 1}{3 - 1}$

Step 3: Calculate the final value.

Average Rate of Change = $\frac{8}{2} = 4$

So, the average rate of change of $f(x) = x^2$ from $x=1$ to $x=3$ is 4.

Part 2: Difference from Instantaneous Rate of Change

The average rate of change and the instantaneous rate of change are related but fundamentally different concepts.

Average Rate of Change:

• It measures how a function changes over an interval (between two points).
• Geometrically, it represents the slope of the secant line that passes through the two points on the function's graph.
• It is an algebraic calculation: $\frac{\Delta y}{\Delta x} = \frac{f(b) - f(a)}{b-a}$.

Instantaneous Rate of Change:

• It measures how a function changes at a single, specific instant or point.
• Geometrically, it represents the slope of the tangent line to the function's graph at that one point.
• It is found using calculus and is the definition of the derivative, $f'(a) = \lim\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$.

In summary, while the average rate of change gives you an overall "speed" across a distance, the instantaneous rate of change gives you the precise "speed" at a particular moment.

The domain of a function is the set of all possible input values (x-values) for which the function is defined and produces a real number as an output. To find the domain of $f(x) = \frac{1}{\sqrt{x-1}}$, we need to identify any values of $x$ that would make the function undefined.

There are two potential issues we must address:

1. The expression inside the square root (the radicand) cannot be negative.
2. The denominator of the fraction cannot be zero.

Given:

The function is $f(x) = \frac{1}{\sqrt{x-1}}$.

To Find:

The domain of the function $f(x)$.

Solution:

We will address the two restrictions one by one and then combine them.

Condition 1: The radicand must be non-negative.

The expression inside the square root, $x-1$, must be greater than or equal to zero for the function to yield a real number.

Solving this inequality for $x$:

$x \geq 1$

This means $x$ must be 1 or any number greater than 1.

Condition 2: The denominator cannot be zero.

The denominator of the fraction is $\sqrt{x-1}$. We must ensure that this does not equal zero.

To find which value of $x$ makes the denominator zero, we can set it equal to zero and solve:

$\sqrt{x-1} = 0$

Squaring both sides:

$x - 1 = 0$

$x = 1$

This means that $x$ cannot be equal to 1.

Combining the Conditions:

From Condition 1, we have $x \geq 1$.

From Condition 2, we have $x \neq 1$.

We need to find the set of numbers that satisfies both conditions. Combining them, we find that $x$ must be strictly greater than 1.

Conclusion:

The domain of the function $f(x) = \frac{1}{\sqrt{x-1}}$ is all real numbers greater than 1.

In interval notation, the domain is $(1, \infty)$.

To evaluate the limit $\lim\limits_{x \to 1} \frac{x^3 - 1}{x - 1}$, we first attempt direct substitution.

Substituting $x=1$ into the expression gives:

$\frac{(1)^3 - 1}{1 - 1} = \frac{1 - 1}{1 - 1} = \frac{0}{0}$

This is an indeterminate form, which indicates that we need to simplify the expression algebraically before evaluating the limit.

Solution using Factorization

We can factor the numerator, $x^3 - 1$, which is a difference of cubes.

The formula for the difference of cubes is $a^3 - b^3 = (a-b)(a^2+ab+b^2)$.

Applying this formula with $a=x$ and $b=1$:

$x^3 - 1^3 = (x-1)(x^2 + x \cdot 1 + 1^2) = (x-1)(x^2+x+1)$

Now, we substitute this factored form back into the limit:

$\lim\limits_{x \to 1} \frac{(x-1)(x^2+x+1)}{x-1}$

Since $x$ is approaching 1 but is not equal to 1, the term $(x-1)$ is not zero, so we can cancel it from the numerator and the denominator:

$\lim\limits_{x \to 1} \frac{\cancel{(x-1)}(x^2+x+1)}{\cancel{x-1}} = \lim\limits_{x \to 1} (x^2+x+1)$

Now, we can evaluate the simplified limit by direct substitution:

$(1)^2 + 1 + 1 = 1 + 1 + 1 = 3$

Alternate Solution using L'Hôpital's Rule

Since the limit is of the indeterminate form $\frac{0}{0}$, we can apply L'Hôpital's Rule. This involves taking the derivative of the numerator and the denominator separately.

Let $f(x) = x^3 - 1$ and $g(x) = x - 1$.

Find their derivatives:

$f'(x) = \frac{d}{dx}(x^3 - 1) = 3x^2$

$g'(x) = \frac{d}{dx}(x - 1) = 1$

According to L'Hôpital's Rule, the original limit is equal to the limit of the ratio of these derivatives:

$\lim\limits_{x \to 1} \frac{x^3 - 1}{x - 1} = \lim\limits_{x \to 1} \frac{f'(x)}{g'(x)} = \lim\limits_{x \to 1} \frac{3x^2}{1}$

Now, we can evaluate this new limit by direct substitution:

$\frac{3(1)^2}{1} = \frac{3}{1} = 3$

Both methods give the same result.

The value of the limit is 3.

To find the derivative of $f(x) = (x^2 + 3x)(x-1)$, we can use the Product Rule, as the function is a product of two simpler functions. An alternative method is to first expand the expression and then differentiate the resulting polynomial term by term.

Given:

The function is $f(x) = (x^2 + 3x)(x-1)$.

To Find:

The derivative of the function, $f'(x)$.

Solution (Using the Product Rule)

The Product Rule for differentiation states that if a function $f(x)$ is the product of two functions, say $u(x)$ and $v(x)$, then its derivative is given by:

$f'(x) = u'(x)v(x) + u(x)v'(x)$

Step 1: Identify the functions $u(x)$ and $v(x)$.

Let $u(x) = x^2 + 3x$

Let $v(x) = x - 1$

Step 2: Find the derivatives of $u(x)$ and $v(x)$.

The derivative of $u(x)$ is:

$u'(x) = \frac{d}{dx}(x^2 + 3x) = 2x + 3$

The derivative of $v(x)$ is:

$v'(x) = \frac{d}{dx}(x - 1) = 1$

Step 3: Apply the Product Rule formula.

$f'(x) = (2x + 3)(x - 1) + (x^2 + 3x)(1)$

Step 4: Simplify the expression.

First, expand the product $(2x + 3)(x - 1)$:

$f'(x) = (2x^2 - 2x + 3x - 3) + (x^2 + 3x)$

$f'(x) = (2x^2 + x - 3) + x^2 + 3x$

Now, combine like terms:

$f'(x) = (2x^2 + x^2) + (x + 3x) - 3$

$f'(x) = 3x^2 + 4x - 3$

Alternate Solution (By Expansion)

Another way to solve this is to first multiply the terms of the function and then differentiate the resulting polynomial.

Step 1: Expand the function $f(x)$.

$f(x) = (x^2 + 3x)(x - 1)$

$f(x) = x^2(x) - x^2(1) + 3x(x) - 3x(1)$

$f(x) = x^3 - x^2 + 3x^2 - 3x$

Combine like terms to get a simplified polynomial:

$f(x) = x^3 + 2x^2 - 3x$

Step 2: Differentiate the expanded polynomial term by term.

Now, we apply the Power Rule to each term:

$f'(x) = \frac{d}{dx}(x^3 + 2x^2 - 3x)$

$f'(x) = \frac{d}{dx}(x^3) + \frac{d}{dx}(2x^2) - \frac{d}{dx}(3x)$

$f'(x) = 3x^2 + 2(2x) - 3$

$f'(x) = 3x^2 + 4x - 3$

Both methods yield the same result. The derivative of $f(x) = (x^2 + 3x)(x-1)$ is $3x^2 + 4x - 3$.

To find the derivative of the function $y = \sqrt{3x^2 + 5}$, we need to use the Chain Rule because it is a composite function (a function inside another function).

The Chain Rule states that if $y = f(u)$ and $u = g(x)$, then the derivative of $y$ with respect to $x$ is:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

Given:

The function is $y = \sqrt{3x^2 + 5}$.

To Find:

The derivative of the function, $\frac{dy}{dx}$.

Solution:

First, it's helpful to rewrite the function using a fractional exponent:

$y = (3x^2 + 5)^{1/2}$

Step 1: Identify the inner function ($u$) and the outer function ($y(u)$).

Let the "inner function" be $u = 3x^2 + 5$.

Then, the "outer function" becomes $y = u^{1/2}$.

Step 2: Find the derivative of the outer function with respect to $u$ ($\frac{dy}{du}$).

Using the Power Rule on $y = u^{1/2}$:

$\frac{dy}{du} = \frac{1}{2}u^{\frac{1}{2}-1} = \frac{1}{2}u^{-1/2}$

This can be rewritten as:

$\frac{dy}{du} = \frac{1}{2\sqrt{u}}$

Step 3: Find the derivative of the inner function with respect to $x$ ($\frac{du}{dx}$).

Differentiating $u = 3x^2 + 5$:

$\frac{du}{dx} = \frac{d}{dx}(3x^2 + 5) = 3(2x) + 0 = 6x$

Step 4: Apply the Chain Rule.

Multiply the results from the previous steps:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

$\frac{dy}{dx} = \left(\frac{1}{2\sqrt{u}}\right) \cdot (6x)$

Step 5: Substitute the expression for $u$ back into the derivative.

Replace $u$ with $3x^2 + 5$:

$\frac{dy}{dx} = \frac{1}{2\sqrt{3x^2 + 5}} \cdot 6x$

Step 6: Simplify the final expression.

$\frac{dy}{dx} = \frac{6x}{2\sqrt{3x^2 + 5}}$

Cancel the common factor of 2 from the numerator and denominator:

$\frac{dy}{dx} = \frac{3x}{\sqrt{3x^2 + 5}}$

Therefore, the derivative of $y = \sqrt{3x^2 + 5}$ is $\frac{dy}{dx} = \frac{3x}{\sqrt{3x^2 + 5}}$.

The Vertical Line Test is a simple graphical method used to determine whether a curve drawn on a coordinate plane represents a function.

The Definition of a Function

Before explaining the test, it's important to remember what a function is. A function is a relation between a set of inputs (the domain) and a set of possible outputs (the range) where each input is related to exactly one output. The key here is that for any given x-value, there can only be one corresponding y-value.

How the Vertical Line Test Works

To apply the Vertical Line Test, follow these steps:

1. Take the graph of the relation in question.
2. Imagine a vertical line (a line of the form $x=c$ for some constant c) that you can move across the graph.
3. Slide this vertical line across the entire graph from left to right.

Interpreting the Results

• If the vertical line intersects the graph at only one point, no matter where you place it, then the graph represents a function. This shows that every x-value in the domain corresponds to exactly one y-value.
• If you can find even one place where the vertical line intersects the graph at more than one point, then the graph does not represent a function. This is because it shows there is at least one x-value that corresponds to multiple y-values, which violates the definition of a function.

Why the Test Works

A vertical line on a graph represents a single, constant x-value. If this line touches the graph at two or more points (e.g., at $(x_1, y_1)$ and $(x_1, y_2)$), it means that the single input value $x_1$ is mapped to two different output values, $y_1$ and $y_2$. This directly violates the rule that each input must have exactly one output.

Examples

1. A Parabola (e.g., $y=x^2$):

An upward-opening parabola will pass the vertical line test. Any vertical line drawn will only ever cross the parabola at a single point. Therefore, $y=x^2$ is a function.

2. A Circle (e.g., $x^2 + y^2 = 9$):

A circle will fail the vertical line test. A vertical line drawn at $x=1$, for instance, will intersect the circle at two points (at $y=\sqrt{8}$ and $y=-\sqrt{8}$). Since one x-value corresponds to two y-values, a circle is not a function.

The range of a function is the set of all possible output values (y-values) that the function can produce for all the x-values in its domain.

Given:

The function is $f(x) = 5 - |x|$.

To Find:

The range of the function $f(x)$.

Solution:

To find the range of $f(x) = 5 - |x|$, we can analyze the behavior of its components. The key component is the absolute value function, $|x|$.

Step 1: Determine the possible values of the absolute value part.

By definition, the absolute value of any real number, $|x|$, is always non-negative. This means its value is always greater than or equal to zero.

Step 2: Build the function $f(x)$ from this inequality.

Our function has the term $-|x|$. To get this term, we can multiply the inequality (i) by -1. When we multiply or divide an inequality by a negative number, we must reverse the direction of the inequality sign.

Multiplying by -1, we get:

Step 3: Add the constant term.

The full function is $5 - |x|$. To get this, we add 5 to both sides of the new inequality:

This simplifies to:

This final inequality tells us that the value of the function $f(x)$ can be 5 or any real number less than 5. The maximum value the function can attain is 5, which occurs when $x=0$. There is no minimum value, as $-|x|$ can become infinitely negative as $x$ moves away from zero.

Conclusion:

The range of the function $f(x) = 5 - |x|$ is the set of all real numbers less than or equal to 5.

In interval notation, the range is $(-\infty, 5]$.

We need to evaluate the limit:

$\lim\limits_{h \to 0} \frac{(x+h)^2 - x^2}{h}$

First, expand the term $(x+h)^2$:

$(x+h)^2 = x^2 + 2xh + h^2$

Substitute this back into the limit expression:

$\lim\limits_{h \to 0} \frac{(x^2 + 2xh + h^2) - x^2}{h}$

Simplify the numerator:

$\lim\limits_{h \to 0} \frac{2xh + h^2}{h}$

Factor out 'h' from the numerator:

$\lim\limits_{h \to 0} \frac{h(2x + h)}{h}$

Cancel out 'h' from the numerator and denominator (since $h \to 0$, $h \neq 0$):

$\lim\limits_{h \to 0} (2x + h)$

Now, substitute $h=0$ into the expression:

$2x + 0 = 2x$

The value of the limit is $2x$.

What does this limit represent?

This limit represents the derivative of the function $f(x) = x^2$ with respect to $x$. It is the definition of the derivative using the first principles.

The derivative of a function at a point gives the instantaneous rate of change of the function at that point, which is also the slope of the tangent line to the function's graph at that point.

For a function $f(x)$ to be continuous at a point $x=c$, the following three conditions must be met:

1. $f(c)$ must be defined.
2. $\lim\limits_{x \to c} f(x)$ must exist.
3. $\lim\limits_{x \to c} f(x) = f(c)$.

In this case, $f(x) = |x|$ and we are checking for continuity at $c=0$. Let's check each condition:

1. Check if $f(0)$ is defined:

$f(0) = |0| = 0$.

So, $f(0)$ is defined and $f(0) = 0$.

2. Check if $\lim\limits_{x \to 0} f(x)$ exists:

For the limit to exist at $x=0$, the left-hand limit and the right-hand limit must be equal.

Left-hand limit: $\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} |x|$

As $x$ approaches 0 from the left side, $x$ is negative. Therefore, $|x| = -x$.

$\lim\limits_{x \to 0^-} |x| = \lim\limits_{x \to 0^-} (-x) = -0 = 0$.

Right-hand limit: $\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} |x|$

As $x$ approaches 0 from the right side, $x$ is positive. Therefore, $|x| = x$.

$\lim\limits_{x \to 0^+} |x| = \lim\limits_{x \to 0^+} (x) = 0$.

Since the left-hand limit equals the right-hand limit ($\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x) = 0$), the limit $\lim\limits_{x \to 0} f(x)$ exists and is equal to 0.

3. Check if $\lim\limits_{x \to 0} f(x) = f(0)$:

From step 1, we found $f(0) = 0$.

From step 2, we found $\lim\limits_{x \to 0} f(x) = 0$.

Since $\lim\limits_{x \to 0} f(x) = f(0)$, the third condition is met.

Conclusion:

All three conditions for continuity are satisfied for $f(x) = |x|$ at $x=0$. Therefore, $f(x) = |x|$ is continuous at $x=0$.

To find the derivative of $y = \frac{x}{x+1}$, we will use the quotient rule. The quotient rule states that if $y = \frac{u}{v}$, then $\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$.

Let $u = x$ and $v = x+1$.

Now, we find the derivatives of $u$ and $v$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(x) = 1$

$\frac{dv}{dx} = \frac{d}{dx}(x+1) = 1$

Now, apply the quotient rule:

$\frac{dy}{dx} = \frac{(x+1)(1) - (x)(1)}{(x+1)^2}$

Simplify the numerator:

$\frac{dy}{dx} = \frac{x + 1 - x}{(x+1)^2}$

$\frac{dy}{dx} = \frac{1}{(x+1)^2}$

Thus, the derivative of $y = \frac{x}{x+1}$ is $\frac{1}{(x+1)^2}$.

To find the derivative of $y = (x^3 - 2x + 5)^{-4}$, we will use the chain rule. The chain rule states that if $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$.

In this case, let $u = x^3 - 2x + 5$ and $y = u^{-4}$.

First, find the derivative of $y$ with respect to $u$ ($f'(u)$):

$\frac{dy}{du} = \frac{d}{du}(u^{-4}) = -4u^{-4-1} = -4u^{-5}$

Next, find the derivative of $u$ with respect to $x$ ($g'(x)$):

$\frac{du}{dx} = \frac{d}{dx}(x^3 - 2x + 5) = 3x^{3-1} - 2(1) + 0 = 3x^2 - 2$

Now, apply the chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

$\frac{dy}{dx} = (-4u^{-5}) \cdot (3x^2 - 2)$

Substitute back $u = x^3 - 2x + 5$:

$\frac{dy}{dx} = -4(x^3 - 2x + 5)^{-5} \cdot (3x^2 - 2)$

We can rewrite this as:

$\frac{dy}{dx} = \frac{-4(3x^2 - 2)}{(x^3 - 2x + 5)^5}$

Alternatively, we can write it as:

$\frac{dy}{dx} = \frac{8 - 12x^2}{(x^3 - 2x + 5)^5}$

The key difference between a relation and a function lies in the uniqueness of the output for each input.

Relation:

A relation is a set of ordered pairs. It describes a relationship between two sets of values (an input set and an output set). For any given input, a relation can have one or more outputs. There are no restrictions on how many times an output value can appear, nor how many output values an input value can have.

For example, if we consider the relation between students and the subjects they are enrolled in, a student can be enrolled in multiple subjects.

Function:

A function is a special type of relation where each input value is associated with exactly one output value. In other words, for every element in the domain (input set), there is precisely one corresponding element in the codomain (output set).

Using the example of students and subjects, if the relation was about a student's primary major, then each student would have only one primary major. This would be a function.

Analogy:

Think of a vending machine:

• Relation: If you press a button (input), it might dispense a drink (output). However, some buttons might be out of order, or a button might dispense different drinks on different tries (not a function).
• Function: In a perfect vending machine that always works, pressing a specific button (input) will always dispense the same, specific item (output). Each button corresponds to only one item.

In summary:

All functions are relations, but not all relations are functions. A function is a relation with the added constraint that each input has only one output.

The function given is $f(x) = 5$ for all $x \in \mathbb{R}$. This is a constant function.

Domain:

The domain of a function is the set of all possible input values ($x$-values) for which the function is defined. In this case, the problem statement explicitly says that the function is defined "for all $x \in \mathbb{R}$". This means that any real number can be used as an input for this function.

Therefore, the domain of $f(x) = 5$ is the set of all real numbers.

Domain: $\mathbb{R}$

In interval notation, this is $(-\infty, \infty)$.

Range:

The range of a function is the set of all possible output values ($f(x)$-values) that the function can produce. For the function $f(x) = 5$, no matter what real number we input for $x$, the output is always $5$.

Therefore, the only output value this function can produce is $5$.

Range: $\{5\}$

We need to evaluate the limit: $\lim\limits_{x \to 1} \frac{x^2 + x - 2}{x - 1}$.

If we try to substitute $x=1$ directly into the expression, we get:

$\frac{(1)^2 + (1) - 2}{1 - 1} = \frac{1 + 1 - 2}{0} = \frac{0}{0}$

This is an indeterminate form, which means we need to simplify the expression before evaluating the limit.

We can factor the numerator, $x^2 + x - 2$. We are looking for two numbers that multiply to -2 and add to 1. These numbers are 2 and -1.

$x^2 + x - 2 = (x + 2)(x - 1)$

Now, substitute the factored numerator back into the limit expression:

$\lim\limits_{x \to 1} \frac{(x + 2)(x - 1)}{x - 1}$

Since $x \to 1$, $x$ is not equal to 1, so $(x - 1)$ is not zero. We can cancel out the $(x - 1)$ term from the numerator and the denominator:

$\lim\limits_{x \to 1} (x + 2)$

Now, we can substitute $x = 1$ into the simplified expression:

$1 + 2 = 3$

Therefore, the value of the limit is 3.

For a function $f(x)$ to be continuous at a point $x=c$, three conditions must be met:

1. $f(c)$ must be defined.
2. $\lim\limits_{x \to c} f(x)$ must exist.
3. $\lim\limits_{x \to c} f(x) = f(c)$.

In this problem, the function is $f(x) = x^2 - 3x + 5$, and we need to check for continuity at $c=0$. This is a polynomial function.

1. Check if $f(0)$ is defined:

$f(0) = (0)^2 - 3(0) + 5 = 0 - 0 + 5 = 5$.

Since $f(0) = 5$, the function is defined at $x=0$.

2. Check if $\lim\limits_{x \to 0} f(x)$ exists:

For a polynomial function, the limit as $x$ approaches a certain value can be found by direct substitution.

$\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} (x^2 - 3x + 5) = (0)^2 - 3(0) + 5 = 0 - 0 + 5 = 5$.

Since the limit exists and is equal to 5, the second condition is met.

3. Check if $\lim\limits_{x \to 0} f(x) = f(0)$:

From step 1, we found $f(0) = 5$.

From step 2, we found $\lim\limits_{x \to 0} f(x) = 5$.

Since $\lim\limits_{x \to 0} f(x) = f(0)$, the third condition for continuity is satisfied.

Conclusion:

All three conditions for continuity are met. Therefore, the function $f(x) = x^2 - 3x + 5$ is continuous at $x=0$. (In fact, all polynomial functions are continuous everywhere).

To find the derivative of $y = (x^2 + 1)^2 (x - 2)$, we need to use the product rule and the chain rule.

The product rule states that if $y = uv$, then $\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$.

Let $u = (x^2 + 1)^2$ and $v = (x - 2)$.

First, find the derivative of $u$ with respect to $x$. We need the chain rule here:

Let $w = x^2 + 1$. Then $u = w^2$.

$\frac{du}{dx} = \frac{du}{dw} \cdot \frac{dw}{dx}$

$\frac{du}{dw} = \frac{d}{dw}(w^2) = 2w$

$\frac{dw}{dx} = \frac{d}{dx}(x^2 + 1) = 2x$

So, $\frac{du}{dx} = (2w) \cdot (2x) = 4xw$. Substituting $w = x^2 + 1$ back:

$\frac{du}{dx} = 4x(x^2 + 1)$

Next, find the derivative of $v$ with respect to $x$:

$\frac{dv}{dx} = \frac{d}{dx}(x - 2) = 1$

Now, apply the product rule: $\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$

$\frac{dy}{dx} = (x^2 + 1)^2 (1) + (x - 2) [4x(x^2 + 1)]$

Now, simplify the expression. We can factor out $(x^2 + 1)$:

$\frac{dy}{dx} = (x^2 + 1) [ (x^2 + 1) + 4x(x - 2) ]$

Expand the terms inside the square brackets:

$\frac{dy}{dx} = (x^2 + 1) [ x^2 + 1 + 4x^2 - 8x ]$

Combine like terms inside the square brackets:

$\frac{dy}{dx} = (x^2 + 1) [ 5x^2 - 8x + 1 ]$

Finally, distribute $(x^2 + 1)$ into the bracketed term:

$\frac{dy}{dx} = x^2(5x^2 - 8x + 1) + 1(5x^2 - 8x + 1)$

$\frac{dy}{dx} = 5x^4 - 8x^3 + x^2 + 5x^2 - 8x + 1$

$\frac{dy}{dx} = 5x^4 - 8x^3 + 6x^2 - 8x + 1$

So, the derivative of $y = (x^2 + 1)^2 (x - 2)$ is $5x^4 - 8x^3 + 6x^2 - 8x + 1$.

The area of a circle is given by the formula $A = \pi r^2$, where $A$ is the area and $r$ is the radius.

We are asked to find the rate of change of the area with respect to its radius. This means we need to find the derivative of the area $A$ with respect to the radius $r$, which is $\frac{dA}{dr}$.

Differentiating $A = \pi r^2$ with respect to $r$:

$\frac{dA}{dr} = \frac{d}{dr}(\pi r^2)$

Since $\pi$ is a constant, we can take it out of the differentiation:

$\frac{dA}{dr} = \pi \frac{d}{dr}(r^2)$

Using the power rule for differentiation ($\frac{d}{dx}(x^n) = nx^{n-1}$):

$\frac{dA}{dr} = \pi (2r^{2-1}) = 2\pi r$

This expression, $2\pi r$, represents the rate of change of the area of a circle with respect to its radius.

We need to find this rate of change when the radius $r = 5$ cm.

Substitute $r=5$ into the expression for $\frac{dA}{dr}$:

$\frac{dA}{dr}\Big|_{r=5} = 2\pi (5)$

$\frac{dA}{dr}\Big|_{r=5} = 10\pi$

The units for the rate of change of area with respect to radius would be the units of area divided by the units of radius, which is cm$^2$/cm, or simply cm.

Therefore, the rate of change of the area of a circle with respect to its radius when $r=5$ cm is $10\pi$ cm.

The domain of a function is the set of all possible input values ($x$-values) for which the function is defined.

For a rational function (a fraction where the numerator and denominator are polynomials), the function is undefined when the denominator is equal to zero.

In this case, the function is $f(x) = \frac{1}{x^2 - 9}$. The denominator is $x^2 - 9$.

To find the values of $x$ that are NOT in the domain, we set the denominator equal to zero:

$x^2 - 9 = 0$

We can solve this equation for $x$. This is a difference of squares, which can be factored as $(x-a)(x+a) = x^2 - a^2$. Here, $a=3$.

$(x - 3)(x + 3) = 0$

For the product of two factors to be zero, at least one of the factors must be zero:

$x - 3 = 0 \implies x = 3$

$x + 3 = 0 \implies x = -3$

So, the function $f(x) = \frac{1}{x^2 - 9}$ is undefined when $x = 3$ or $x = -3$. These are the values that must be excluded from the domain.

The domain of the function is the set of all real numbers except for $3$ and $-3$.

Domain: $\{x \in \mathbb{R} \mid x \neq 3 \text{ and } x \neq -3\}$

In interval notation, the domain is:

$(-\infty, -3) \cup (-3, 3) \cup (3, \infty)$

We need to evaluate the limit: $\lim\limits_{x \to 0} \frac{(x+2)^3 - 8}{x}$.

If we substitute $x=0$ directly, we get $\frac{(0+2)^3 - 8}{0} = \frac{2^3 - 8}{0} = \frac{8 - 8}{0} = \frac{0}{0}$. This is an indeterminate form.

We can solve this by expanding the term $(x+2)^3$ or by using the definition of the derivative.

Method 1: Expansion

Expand $(x+2)^3$ using the binomial theorem or by direct multiplication:

$(x+2)^3 = x^3 + 3(x^2)(2) + 3(x)(2^2) + 2^3$

$(x+2)^3 = x^3 + 6x^2 + 12x + 8$

Substitute this expansion back into the limit expression:

$\lim\limits_{x \to 0} \frac{(x^3 + 6x^2 + 12x + 8) - 8}{x}$

Simplify the numerator:

$\lim\limits_{x \to 0} \frac{x^3 + 6x^2 + 12x}{x}$

Factor out $x$ from the numerator:

$\lim\limits_{x \to 0} \frac{x(x^2 + 6x + 12)}{x}$

Since $x \to 0$, $x \neq 0$, so we can cancel out the $x$ terms:

$\lim\limits_{x \to 0} (x^2 + 6x + 12)$

Now, substitute $x=0$ into the simplified expression:

$(0)^2 + 6(0) + 12 = 0 + 0 + 12 = 12$

Method 2: Using the definition of the derivative

Recall the definition of the derivative of a function $f(a)$ at a point $a$: $f'(a) = \lim\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$.

Let $f(x) = x^3$. We want to evaluate the derivative of $f(x)$ at $x=2$.

Let $g(x) = (x+2)^3$. We are interested in the limit as $x \to 0$. Let $u = x+2$. As $x \to 0$, $u \to 2$. So the expression becomes:

$\lim\limits_{u \to 2} \frac{u^3 - 2^3}{u - 2}$

This is the definition of the derivative of the function $f(u) = u^3$ at $u=2$.

The derivative of $f(u) = u^3$ is $f'(u) = 3u^2$.

Evaluating this derivative at $u=2$: $f'(2) = 3(2)^2 = 3(4) = 12$.

Both methods yield the same result.

The value of the limit is 12.

The definition of the derivative of a function $f(x)$ is given by:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Given the function $f(x) = x^2$.

First, find $f(x+h)$:

$f(x+h) = (x+h)^2$

Expand $(x+h)^2$:

$f(x+h) = x^2 + 2xh + h^2$

Now, substitute $f(x+h)$ and $f(x)$ into the definition of the derivative:

$f'(x) = \lim\limits_{h \to 0} \frac{(x^2 + 2xh + h^2) - x^2}{h}$

Simplify the numerator:

$f'(x) = \lim\limits_{h \to 0} \frac{2xh + h^2}{h}$

Factor out $h$ from the numerator:

$f'(x) = \lim\limits_{h \to 0} \frac{h(2x + h)}{h}$

Since $h \to 0$, $h \neq 0$, we can cancel out the $h$ terms:

$f'(x) = \lim\limits_{h \to 0} (2x + h)$

Now, substitute $h=0$ into the expression:

$f'(x) = 2x + 0$

$f'(x) = 2x$

Therefore, the derivative of $f(x) = x^2$ using the definition is $f'(x) = 2x$.

First, rewrite the function using exponent notation:

$y = (x^2 + 5x)^{1/3}$

To find the derivative of this function, we will use the chain rule. The chain rule states that if $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$.

Let $u = x^2 + 5x$ and $y = u^{1/3}$.

Find the derivative of $y$ with respect to $u$ ($f'(u)$):

$\frac{dy}{du} = \frac{d}{du}(u^{1/3})$

Using the power rule, $\frac{d}{du}(u^n) = nu^{n-1}$:

$\frac{dy}{du} = \frac{1}{3}u^{\frac{1}{3}-1} = \frac{1}{3}u^{-\frac{2}{3}}$

Find the derivative of $u$ with respect to $x$ ($g'(x)$):

$\frac{du}{dx} = \frac{d}{dx}(x^2 + 5x)$

Using the power rule and the constant multiple rule:

$\frac{du}{dx} = 2x^{2-1} + 5(1) = 2x + 5$

Now, apply the chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

$\frac{dy}{dx} = \left(\frac{1}{3}u^{-\frac{2}{3}}\right) \cdot (2x + 5)$

Substitute back $u = x^2 + 5x$:

$\frac{dy}{dx} = \frac{1}{3}(x^2 + 5x)^{-\frac{2}{3}} (2x + 5)$

We can rewrite this with a positive exponent and in fractional form:

$\frac{dy}{dx} = \frac{2x + 5}{3(x^2 + 5x)^{\frac{2}{3}}}$

This can also be written using radical notation:

$\frac{dy}{dx} = \frac{2x + 5}{3\sqrt[3]{(x^2 + 5x)^2}}$

Thus, the derivative of $y = \sqrt[3]{x^2 + 5x}$ is $\frac{2x + 5}{3\sqrt[3]{(x^2 + 5x)^2}}$.

The function given is $f(x) = x$. This is a linear function, often referred to as the identity function.

Graphing the function:

To sketch the graph of $f(x) = x$, we can identify a few key points:

• When $x=0$, $f(x) = 0$. So, the point (0, 0) is on the graph (this is the origin).
• When $x=1$, $f(x) = 1$. So, the point (1, 1) is on the graph.
• When $x=-1$, $f(x) = -1$. So, the point (-1, -1) is on the graph.

The graph of $f(x) = x$ is a straight line that passes through the origin with a slope of 1. It makes an angle of 45 degrees with the positive x-axis.

*(A visual representation of the graph would typically be shown here with axes, the origin, and the line $y=x$ passing through it.)*

Domain:

The domain of a function is the set of all possible input values ($x$-values). For $f(x) = x$, there are no restrictions on the value of $x$. We can input any real number for $x$, and the function will produce a real number output.

Domain: $\mathbb{R}$ (all real numbers)

In interval notation: $(-\infty, \infty)$

Range:

The range of a function is the set of all possible output values ($f(x)$-values). Since the graph of $f(x) = x$ extends infinitely upwards and downwards through all real numbers, any real number can be an output.

Range: $\mathbb{R}$ (all real numbers)

In interval notation: $(-\infty, \infty)$

The domain of a rational function is the set of all real numbers except for the values of $x$ that make the denominator equal to zero.

The given function is $f(x) = \frac{x+2}{x^2 - 4}$.

The denominator is $x^2 - 4$.

To find the values of $x$ that are NOT in the domain, we set the denominator equal to zero:

$x^2 - 4 = 0$

This is a difference of squares, which can be factored as:

$(x - 2)(x + 2) = 0$

Setting each factor to zero gives us the values of $x$ that make the denominator zero:

$x - 2 = 0 \implies x = 2$

$x + 2 = 0 \implies x = -2$

Therefore, the function $f(x)$ is undefined when $x=2$ or $x=-2$. These values must be excluded from the domain.

The domain of $f(x)$ is the set of all real numbers except $2$ and $-2$.

Domain: $\{x \in \mathbb{R} \mid x \neq 2 \text{ and } x \neq -2\}$

In interval notation, the domain is:

$(-\infty, -2) \cup (-2, 2) \cup (2, \infty)$

We need to evaluate the limit: $\lim\limits_{x \to -1} \frac{x^2 - 1}{x+1}$.

If we try to substitute $x = -1$ directly into the expression, we get:

$\frac{(-1)^2 - 1}{-1 + 1} = \frac{1 - 1}{0} = \frac{0}{0}$

This is an indeterminate form, which means we need to simplify the expression.

The numerator is a difference of squares, $x^2 - 1^2$, which can be factored as $(x - 1)(x + 1)$.

Substitute the factored numerator back into the limit expression:

$\lim\limits_{x \to -1} \frac{(x - 1)(x + 1)}{x + 1}$

Since $x \to -1$, $x$ is not equal to $-1$, so $(x + 1)$ is not zero. We can cancel out the $(x + 1)$ term from the numerator and the denominator:

$\lim\limits_{x \to -1} (x - 1)$

Now, substitute $x = -1$ into the simplified expression:

$(-1) - 1 = -2$

Therefore, the value of the limit is -2.

For the function $f(x)$ to be continuous at $x=1$, the following three conditions must be met:

1. $f(1)$ must be defined.
2. $\lim\limits_{x \to 1} f(x)$ must exist.
3. $\lim\limits_{x \to 1} f(x) = f(1)$.

Let's analyze each condition for the given piecewise function:

1. $f(1)$ must be defined:

The function is defined as $f(x) = ax+5$ for $x \leq 1$. So, for $x=1$, we use the first part of the definition:

$f(1) = a(1) + 5 = a + 5$.

So, $f(1)$ is defined as $a+5$.

2. $\lim\limits_{x \to 1} f(x)$ must exist:

For the limit to exist at $x=1$, the left-hand limit and the right-hand limit must be equal.

Left-hand limit: As $x$ approaches 1 from the left ($x \leq 1$), we use the first part of the function:

$\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^-} (ax+5) = a(1) + 5 = a+5$.

Right-hand limit: As $x$ approaches 1 from the right ($x > 1$), we use the second part of the function:

$\lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1^+} (x-2) = 1-2 = -1$.

For the limit to exist, the left-hand limit must equal the right-hand limit:

Solving for $a$ from equation (i):

$a = -1 - 5$

$a = -6$

3. $\lim\limits_{x \to 1} f(x) = f(1)$:

From step 1, $f(1) = a+5$.

From step 2, if the limit exists, it is equal to $a+5$ (from the left-hand limit) and $-1$ (from the right-hand limit). For continuity, these must be equal.

So, we need $a+5 = -1$. This is the same equation we derived in step 2.

Solving $a+5 = -1$ for $a$, we get $a = -6$.

Therefore, the value of $a$ for which the function is continuous at $x=1$ is $-6$.

To find the derivative of $y = \frac{x^2 - 1}{x^2 + 1}$, we will use the quotient rule. The quotient rule states that if $y = \frac{u}{v}$, then $\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$.

Let $u = x^2 - 1$ and $v = x^2 + 1$.

Now, find the derivatives of $u$ and $v$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(x^2 - 1) = 2x$

$\frac{dv}{dx} = \frac{d}{dx}(x^2 + 1) = 2x$

Apply the quotient rule:

$\frac{dy}{dx} = \frac{(x^2 + 1)(2x) - (x^2 - 1)(2x)}{(x^2 + 1)^2}$

Now, simplify the numerator. We can factor out $2x$ from both terms:

$\frac{dy}{dx} = \frac{2x[(x^2 + 1) - (x^2 - 1)]}{(x^2 + 1)^2}$

Simplify the expression inside the square brackets:

$(x^2 + 1) - (x^2 - 1) = x^2 + 1 - x^2 + 1 = 2$

Substitute this back into the derivative expression:

$\frac{dy}{dx} = \frac{2x(2)}{(x^2 + 1)^2}$

$\frac{dy}{dx} = \frac{4x}{(x^2 + 1)^2}$

Thus, the derivative of $y = \frac{x^2 - 1}{x^2 + 1}$ is $\frac{4x}{(x^2 + 1)^2}$.

First, rewrite the function using exponent notation:

$y = (x^2 + 4)^{-1/2}$

To find the derivative of this function, we will use the chain rule. The chain rule states that if $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$.

Let $u = x^2 + 4$ and $y = u^{-1/2}$.

Find the derivative of $y$ with respect to $u$ ($f'(u)$):

$\frac{dy}{du} = \frac{d}{du}(u^{-1/2})$

Using the power rule, $\frac{d}{du}(u^n) = nu^{n-1}$:

$\frac{dy}{du} = -\frac{1}{2}u^{-\frac{1}{2}-1} = -\frac{1}{2}u^{-\frac{3}{2}}$

Find the derivative of $u$ with respect to $x$ ($g'(x)$):

$\frac{du}{dx} = \frac{d}{dx}(x^2 + 4)$

Using the power rule and the constant rule:

$\frac{du}{dx} = 2x^{2-1} + 0 = 2x$

Now, apply the chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

$\frac{dy}{dx} = \left(-\frac{1}{2}u^{-\frac{3}{2}}\right) \cdot (2x)$

Substitute back $u = x^2 + 4$:

$\frac{dy}{dx} = -\frac{1}{2}(x^2 + 4)^{-\frac{3}{2}} (2x)$

Simplify the expression:

$\frac{dy}{dx} = -x(x^2 + 4)^{-\frac{3}{2}}$

We can rewrite this with a positive exponent and in fractional form:

$\frac{dy}{dx} = -\frac{x}{(x^2 + 4)^{\frac{3}{2}}}$

This can also be written using radical notation:

$\frac{dy}{dx} = -\frac{x}{\sqrt{(x^2 + 4)^3}}$

Thus, the derivative of $y = \frac{1}{\sqrt{x^2 + 4}}$ is $-\frac{x}{(x^2 + 4)^{3/2}}$.

The function given is $f(x) = \frac{1}{x}$.

Domain:

The domain of a function is the set of all possible input values ($x$-values) for which the function is defined. For the function $f(x) = \frac{1}{x}$, the function is undefined when the denominator is zero.

Setting the denominator to zero:

$x = 0$

Therefore, $x=0$ must be excluded from the domain.

The domain of $f(x) = \frac{1}{x}$ is the set of all real numbers except 0.

Domain: $\{x \in \mathbb{R} \mid x \neq 0\}$

In interval notation: $(-\infty, 0) \cup (0, \infty)$

Range:

The range of a function is the set of all possible output values ($f(x)$-values). For $f(x) = \frac{1}{x}$, we want to know what values $y = \frac{1}{x}$ can take.

If we try to solve for $x$ in terms of $y$, we get $x = \frac{1}{y}$.

For $x$ to be a real number, the denominator $y$ cannot be zero.

Therefore, $y$ can be any real number except 0.

The range of $f(x) = \frac{1}{x}$ is the set of all real numbers except 0.

Range: $\{y \in \mathbb{R} \mid y \neq 0\}$

In interval notation: $(-\infty, 0) \cup (0, \infty)$

The graph of $f(x) = 1/x$ is a hyperbola with two branches, one in the first quadrant and one in the third quadrant, with the x-axis and y-axis as its asymptotes.

We need to evaluate the limit: $\lim\limits_{x \to 4} \frac{x^2 - 16}{\sqrt{x} - 2}$.

If we substitute $x=4$ directly into the expression, we get:

$\frac{(4)^2 - 16}{\sqrt{4} - 2} = \frac{16 - 16}{2 - 2} = \frac{0}{0}$

This is an indeterminate form, so we need to simplify the expression. We can do this by rationalizing the denominator.

To rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{x} + 2$:

$\lim\limits_{x \to 4} \frac{x^2 - 16}{\sqrt{x} - 2} \times \frac{\sqrt{x} + 2}{\sqrt{x} + 2}$

Multiply the numerators:

$(x^2 - 16)(\sqrt{x} + 2)$

Multiply the denominators (using $(a-b)(a+b) = a^2 - b^2$):

$(\sqrt{x} - 2)(\sqrt{x} + 2) = (\sqrt{x})^2 - (2)^2 = x - 4$

So the expression becomes:

$\lim\limits_{x \to 4} \frac{(x^2 - 16)(\sqrt{x} + 2)}{x - 4}$

Notice that the term $x^2 - 16$ can be factored as a difference of squares: $x^2 - 16 = (x - 4)(x + 4)$.

Substitute this factorization back into the expression:

$\lim\limits_{x \to 4} \frac{(x - 4)(x + 4)(\sqrt{x} + 2)}{x - 4}$

Since $x \to 4$, $x$ is not equal to 4, so $(x - 4)$ is not zero. We can cancel out the $(x - 4)$ terms:

$\lim\limits_{x \to 4} (x + 4)(\sqrt{x} + 2)$

Now, substitute $x = 4$ into the simplified expression:

$(4 + 4)(\sqrt{4} + 2)$

$= (8)(2 + 2)$

$= (8)(4)$

$= 32$

Therefore, the value of the limit is 32.

The geometrical interpretation of the derivative of a function $f(x)$ at a specific point $x=a$ is the slope of the tangent line to the graph of the function at that point.

Let's break this down:

1. Tangent Line:

A tangent line to a curve at a point is a straight line that "just touches" the curve at that single point. It has the same direction as the curve at that point.

2. Slope:

The slope of a line is a measure of its steepness. It tells us how much the $y$-value changes for a unit change in the $x$-value (rise over run).

3. Derivative as a Limit of Secant Slopes:

The derivative is formally defined as the limit of the slope of secant lines. A secant line passes through two points on the curve. As these two points get infinitely close to each other, the secant line approaches the tangent line.

Mathematically, the derivative $f'(a)$ is given by:

$f'(a) = \lim\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$

Here, $\frac{f(a+h) - f(a)}{h}$ represents the slope of the secant line passing through points $(a, f(a))$ and $(a+h, f(a+h))$. As $h$ approaches 0, these two points merge into a single point $(a, f(a))$, and the slope of the secant line becomes the slope of the tangent line.

In summary:

• If $f'(a) > 0$, the tangent line has a positive slope, meaning the function is increasing at $x=a$.
• If $f'(a) < 0$, the tangent line has a negative slope, meaning the function is decreasing at $x=a$.
• If $f'(a) = 0$, the tangent line is horizontal, meaning the function has a stationary point (like a local maximum, local minimum, or a saddle point) at $x=a$.

The derivative at a point tells us the instantaneous rate of change of the function's output with respect to its input at that precise point, and geometrically, this rate of change is represented by the steepness and direction of the tangent line at that point on the curve.

To find the derivative of $y = (x^4 - 3x^2 + 2)^6$, we will use the chain rule. The chain rule states that if $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$.

Let $u = x^4 - 3x^2 + 2$ and $y = u^6$.

First, find the derivative of $y$ with respect to $u$ ($f'(u)$):

$\frac{dy}{du} = \frac{d}{du}(u^6)$

Using the power rule, $\frac{d}{du}(u^n) = nu^{n-1}$:

$\frac{dy}{du} = 6u^{6-1} = 6u^5$

Next, find the derivative of $u$ with respect to $x$ ($g'(x)$):

$\frac{du}{dx} = \frac{d}{dx}(x^4 - 3x^2 + 2)$

Using the power rule and the constant multiple rule:

$\frac{du}{dx} = 4x^{4-1} - 3(2x^{2-1}) + 0 = 4x^3 - 6x$

Now, apply the chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

$\frac{dy}{dx} = (6u^5) \cdot (4x^3 - 6x)$

Substitute back $u = x^4 - 3x^2 + 2$:

$\frac{dy}{dx} = 6(x^4 - 3x^2 + 2)^5 (4x^3 - 6x)$

We can factor out $2x$ from the second part for a slightly cleaner form:

$\frac{dy}{dx} = 6(x^4 - 3x^2 + 2)^5 \cdot 2x(2x^2 - 3)$

$\frac{dy}{dx} = 12x(2x^2 - 3)(x^4 - 3x^2 + 2)^5$

Thus, the derivative of $(x^4 - 3x^2 + 2)^6$ with respect to $x$ is $6(x^4 - 3x^2 + 2)^5 (4x^3 - 6x)$ or $12x(2x^2 - 3)(x^4 - 3x^2 + 2)^5$.

For the function $f(x) = \frac{1}{\sqrt{4 - x^2}}$ to be defined, two conditions must be met:

1. The expression under the square root must be non-negative (greater than or equal to zero).
2. The denominator cannot be zero.

Combining these two conditions, the expression under the square root must be strictly positive (greater than zero).

So, we must have: $4 - x^2 > 0$.

To solve this inequality, we first find the values of $x$ for which $4 - x^2 = 0$.

$4 - x^2 = 0$

$4 = x^2$

$x = \pm \sqrt{4}$

$x = \pm 2$

These values, $x=-2$ and $x=2$, divide the number line into three intervals: $(-\infty, -2)$, $(-2, 2)$, and $(2, \infty)$. We need to test a value from each interval to see where the inequality $4 - x^2 > 0$ holds true.

Test interval $(-\infty, -2)$: Let's choose $x = -3$.

$4 - (-3)^2 = 4 - 9 = -5$. Since $-5 \ngtr 0$, this interval is not part of the domain.

Test interval $(-2, 2)$: Let's choose $x = 0$.

$4 - (0)^2 = 4 - 0 = 4$. Since $4 > 0$, this interval is part of the domain.

Test interval $(2, \infty)$: Let's choose $x = 3$.

$4 - (3)^2 = 4 - 9 = -5$. Since $-5 \ngtr 0$, this interval is not part of the domain.

The inequality $4 - x^2 > 0$ is satisfied only for the interval $(-2, 2)$.

Therefore, the domain of the function $f(x) = \frac{1}{\sqrt{4 - x^2}}$ is $(-2, 2)$.

We need to evaluate the limit: $\lim\limits_{h \to 0} \frac{(x+h)^3 - x^3}{h}$ for a fixed $x$. *(Note: The question states $\lim\limits_{x \to 0}$, but the expression clearly involves $h$ approaching 0, and $x$ is fixed. We will proceed with $\lim\limits_{h \to 0}$ as it aligns with the structure of the expression and the concept of derivative definition.)*

First, let's expand the term $(x+h)^3$ using the binomial expansion formula $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$.

Let $a = x$ and $b = h$.

$(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$

Now, substitute this expansion back into the limit expression:

$\lim\limits_{h \to 0} \frac{(x^3 + 3x^2h + 3xh^2 + h^3) - x^3}{h}$

Simplify the numerator by cancelling out the $x^3$ terms:

$\lim\limits_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h}$

Factor out $h$ from each term in the numerator:

$\lim\limits_{h \to 0} \frac{h(3x^2 + 3xh + h^2)}{h}$

Since $h \to 0$, $h$ is not equal to 0. Therefore, we can cancel out the $h$ in the numerator and the denominator:

$\lim\limits_{h \to 0} (3x^2 + 3xh + h^2)$

Now, we can substitute $h=0$ into the simplified expression, as the expression is no longer indeterminate:

$3x^2 + 3x(0) + (0)^2 = 3x^2 + 0 + 0 = 3x^2$

This limit represents the derivative of the function $f(x) = x^3$ with respect to $x$, evaluated using the definition of the derivative.

Therefore, the value of the limit is $3x^2$.

The given function is $f(x) = \frac{x^2 - 1}{x - 1}$.

Domain:

The domain of a rational function is the set of all real numbers for which the denominator is not zero. The denominator is $x-1$.

Set the denominator to zero:

$x - 1 = 0 \implies x = 1$

Thus, $x=1$ is excluded from the domain.

Domain: $\{x \in \mathbb{R} \mid x \neq 1\}$

In interval notation: $(-\infty, 1) \cup (1, \infty)$

Range:

We can simplify the function by factoring the numerator:

$x^2 - 1 = (x - 1)(x + 1)$

So, $f(x) = \frac{(x - 1)(x + 1)}{x - 1}$.

For $x \neq 1$, we can cancel out the $(x - 1)$ terms:

$f(x) = x + 1$, for $x \neq 1$.

This means the graph of $f(x)$ is the line $y = x+1$, but with a hole at $x=1$.

To find the value of $y$ where the hole occurs, substitute $x=1$ into the simplified expression $y = x+1$:

$y = 1 + 1 = 2$

So, the function's graph is the line $y=x+1$ excluding the point $(1, 2)$.

The range is all real numbers except for the value $2$.

Range: $\{y \in \mathbb{R} \mid y \neq 2\}$

In interval notation: $(-\infty, 2) \cup (2, \infty)$

Discontinuity at $x=1$:

Since the function can be simplified by canceling a factor of $(x-1)$ from the numerator and denominator, and this factor is zero at $x=1$, the discontinuity is a removable discontinuity (also called a hole).

To confirm, let's check the conditions for continuity:

• $f(1)$ is undefined (from the domain analysis).
• The limit as $x \to 1$: $\lim\limits_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim\limits_{x \to 1} (x+1) = 1+1 = 2$. So, the limit exists.

Because $f(1)$ is undefined, the function is discontinuous at $x=1$. Since the limit exists, it is a removable discontinuity.

The given function is $f(x) = \sqrt{9 - x^2}$.

Domain:

For the function to be defined, the expression under the square root must be non-negative (greater than or equal to zero).

$9 - x^2 \geq 0$

Add $x^2$ to both sides:

$9 \geq x^2$

This inequality means that $x^2$ must be less than or equal to 9. Taking the square root of both sides (and considering both positive and negative roots):

$|x| \leq 3$

This implies that $-3 \leq x \leq 3$.

Domain: $\{x \in \mathbb{R} \mid -3 \leq x \leq 3\}$

In interval notation: $[-3, 3]$

Range:

The function involves a square root, which by convention gives the principal (non-negative) root. Therefore, the output $f(x)$ will always be greater than or equal to zero.

The minimum value of $f(x)$ occurs when $9 - x^2$ is minimum within the domain, which is 0 (at $x=-3$ or $x=3$). In this case, $f(x) = \sqrt{0} = 0$.

The maximum value of $f(x)$ occurs when $9 - x^2$ is maximum, which happens when $x=0$. In this case, $f(x) = \sqrt{9 - 0^2} = \sqrt{9} = 3$.

So, the output values range from 0 to 3.

Range: $\{y \in \mathbb{R} \mid 0 \leq y \leq 3\}$

In interval notation: $[0, 3]$

Graphing the function:

To understand the graph, let's square both sides of the equation $y = \sqrt{9 - x^2}$ (remembering that $y \geq 0$):

$y^2 = 9 - x^2$

Rearranging the terms, we get:

$x^2 + y^2 = 9$

This is the equation of a circle centered at the origin $(0,0)$ with a radius $r$, where $r^2 = 9$, so $r = 3$.

However, our original function $f(x) = \sqrt{9 - x^2}$ includes the square root, which means $y$ must be non-negative ($y \geq 0$).

Therefore, the graph of $f(x) = \sqrt{9 - x^2}$ is the upper semi-circle of a circle centered at the origin with radius 3.

*(A visual representation of the graph would show a semi-circle in the upper half of the xy-plane, with the x-axis from -3 to 3 as its base and the highest point at (0, 3).)*

We need to evaluate the limit: $\lim\limits_{x \to 3} \frac{x^4 - 81}{x^3 - 27}$.

If we substitute $x=3$ directly into the expression, we get:

$\frac{3^4 - 81}{3^3 - 27} = \frac{81 - 81}{27 - 27} = \frac{0}{0}$

This is an indeterminate form, which means we need to simplify the expression. We can do this by factoring both the numerator and the denominator.

Factoring the numerator: $x^4 - 81$ is a difference of squares ($a^2 - b^2 = (a-b)(a+b)$) where $a = x^2$ and $b = 9$.

$x^4 - 81 = (x^2)^2 - 9^2 = (x^2 - 9)(x^2 + 9)$

The term $(x^2 - 9)$ is also a difference of squares, where $a=x$ and $b=3$:

$x^2 - 9 = (x - 3)(x + 3)$

So, the factored numerator is: $(x - 3)(x + 3)(x^2 + 9)$

Factoring the denominator: $x^3 - 27$ is a difference of cubes ($a^3 - b^3 = (a-b)(a^2 + ab + b^2)$) where $a=x$ and $b=3$.

$x^3 - 27 = x^3 - 3^3 = (x - 3)(x^2 + 3x + 3^2) = (x - 3)(x^2 + 3x + 9)$

Now, substitute the factored forms back into the limit expression:

$\lim\limits_{x \to 3} \frac{(x - 3)(x + 3)(x^2 + 9)}{(x - 3)(x^2 + 3x + 9)}$

Since $x \to 3$, $x \neq 3$, so $(x - 3)$ is not zero. We can cancel out the $(x - 3)$ term:

$\lim\limits_{x \to 3} \frac{(x + 3)(x^2 + 9)}{x^2 + 3x + 9}$

Now, substitute $x = 3$ into the simplified expression:

$\frac{(3 + 3)(3^2 + 9)}{3^2 + 3(3) + 9}$

= $\frac{(6)(9 + 9)}{9 + 9 + 9}$

= $\frac{(6)(18)}{27}$

= $\frac{108}{27}$

To simplify $\frac{108}{27}$, we can notice that $108 = 4 \times 27$.

= 4

Therefore, the value of the limit is 4.

Part 1: Find the derivative of $f(x) = \frac{1}{x}$ using the definition.

The definition of the derivative is $f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$.

Given $f(x) = \frac{1}{x}$.

First, find $f(x+h)$: $f(x+h) = \frac{1}{x+h}$.

Substitute $f(x+h)$ and $f(x)$ into the definition:

$f'(x) = \lim\limits_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h}$

To simplify the numerator, find a common denominator:

$f'(x) = \lim\limits_{h \to 0} \frac{\frac{x - (x+h)}{x(x+h)}}{h}$

$f'(x) = \lim\limits_{h \to 0} \frac{\frac{x - x - h}{x(x+h)}}{h}$

$f'(x) = \lim\limits_{h \to 0} \frac{\frac{-h}{x(x+h)}}{h}$

Now, divide by $h$ (which is the same as multiplying by $\frac{1}{h}$):

$f'(x) = \lim\limits_{h \to 0} \frac{-h}{x(x+h)} \cdot \frac{1}{h}$

Since $h \to 0$, $h \neq 0$, we can cancel out the $h$ terms:

$f'(x) = \lim\limits_{h \to 0} \frac{-1}{x(x+h)}$

Now, substitute $h=0$ into the simplified expression:

$f'(x) = \frac{-1}{x(x+0)}$

$f'(x) = \frac{-1}{x^2}$

So, the derivative of $f(x) = \frac{1}{x}$ is $f'(x) = -\frac{1}{x^2}$.

Part 2: Find the equation of the tangent line to the curve $y = 1/x$ at $x=2$.

To find the equation of a line, we need a point on the line and its slope. We can use the point-slope form of a linear equation: $y - y_1 = m(x - x_1)$.

Find the point $(x_1, y_1)$:

We are given $x_1 = 2$. To find $y_1$, substitute $x=2$ into the original function $y = 1/x$.

$y_1 = \frac{1}{2}$

So, the point is $(2, \frac{1}{2})$.

Find the slope ($m$):

The slope of the tangent line at a point is the value of the derivative at that point.

We found the derivative to be $f'(x) = -\frac{1}{x^2}$.

Now, evaluate the derivative at $x=2$ to find the slope:

$m = f'(2) = -\frac{1}{(2)^2} = -\frac{1}{4}$

Write the equation of the tangent line:

Using the point-slope form $y - y_1 = m(x - x_1)$ with $(x_1, y_1) = (2, \frac{1}{2})$ and $m = -\frac{1}{4}$:

$y - \frac{1}{2} = -\frac{1}{4}(x - 2)$

Now, we can simplify this equation to the slope-intercept form ($y = mx + b$):

$y - \frac{1}{2} = -\frac{1}{4}x + (-\frac{1}{4})(-2)$

$y - \frac{1}{2} = -\frac{1}{4}x + \frac{2}{4}$

$y - \frac{1}{2} = -\frac{1}{4}x + \frac{1}{2}$

Add $\frac{1}{2}$ to both sides:

$y = -\frac{1}{4}x + \frac{1}{2} + \frac{1}{2}$

$y = -\frac{1}{4}x + 1$

The equation of the tangent line to the curve $y = 1/x$ at $x=2$ is $y = -\frac{1}{4}x + 1$.

For a function $f(x)$ to be continuous at a point $x=c$, the following three conditions must be met:

1. $f(c)$ must be defined.
2. $\lim\limits_{x \to c} f(x)$ must exist.
3. $\lim\limits_{x \to c} f(x) = f(c)$.

In this problem, we check for continuity at $c=1$. The function is defined piecewise:

$f(x) = \begin{cases} x^2 + 3 & , & x \leq 1 \\ 5 - x^2 & , & x > 1 \end{cases}$

1. Check if $f(1)$ is defined:

Since $x \leq 1$ includes $x=1$, we use the first part of the definition:

$f(1) = (1)^2 + 3 = 1 + 3 = 4$.

So, $f(1)$ is defined and $f(1) = 4$.

2. Check if $\lim\limits_{x \to 1} f(x)$ exists:

We need to evaluate the left-hand limit and the right-hand limit.

Left-hand limit: As $x$ approaches 1 from the left ($x \leq 1$), we use $f(x) = x^2 + 3$.

$\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^-} (x^2 + 3) = (1)^2 + 3 = 1 + 3 = 4$.

Right-hand limit: As $x$ approaches 1 from the right ($x > 1$), we use $f(x) = 5 - x^2$.

$\lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1^+} (5 - x^2) = 5 - (1)^2 = 5 - 1 = 4$.

Since the left-hand limit and the right-hand limit are equal ($4 = 4$), the limit $\lim\limits_{x \to 1} f(x)$ exists and is equal to 4.

3. Check if $\lim\limits_{x \to 1} f(x) = f(1)$:

From step 1, $f(1) = 4$.

From step 2, $\lim\limits_{x \to 1} f(x) = 4$.

Since $\lim\limits_{x \to 1} f(x) = f(1)$, the third condition for continuity is satisfied.

Conclusion on Continuity:

All three conditions for continuity are met. Therefore, the function $f(x)$ is continuous at $x=1$.

Sketching the graph:

The function consists of two parts:

• For $x \leq 1$, $f(x) = x^2 + 3$. This is a parabola opening upwards, shifted 3 units up. At $x=1$, $f(1) = 1^2 + 3 = 4$. This part includes the point $(1, 4)$ as a solid dot.
• For $x > 1$, $f(x) = 5 - x^2$. This is a parabola opening downwards, shifted 5 units up. At $x=1$, the value would be $5 - 1^2 = 4$. Since $x > 1$, this part of the graph starts just after $x=1$. The point $(1, 4)$ would be an open circle if the limit from the right did not match $f(1)$, but since it does match, the function is continuous.

The graph will be a continuous curve passing through $(1, 4)$.

*(A visual representation would show the left part of a parabola $y=x^2+3$ up to $x=1$, and the right part of a parabola $y=5-x^2$ for $x>1$. Both pieces meet at the point $(1, 4)$.)*

Part 1: Find the derivative using the product rule.

The product rule states that if $y = uv$, then $\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$.

Let $u = x^2 + 5$ and $v = 2x^3 - x^2 + 1$.

Find the derivatives of $u$ and $v$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(x^2 + 5) = 2x$

$\frac{dv}{dx} = \frac{d}{dx}(2x^3 - x^2 + 1) = 2(3x^2) - 2x + 0 = 6x^2 - 2x$

Apply the product rule:

$\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$

$\frac{dy}{dx} = (x^2 + 5)(6x^2 - 2x) + (2x^3 - x^2 + 1)(2x)$

Expand the terms:

$(x^2 + 5)(6x^2 - 2x) = x^2(6x^2 - 2x) + 5(6x^2 - 2x)$

= $6x^4 - 2x^3 + 30x^2 - 10x$

$(2x^3 - x^2 + 1)(2x) = 2x(2x^3) - 2x(x^2) + 2x(1)$

= $4x^4 - 2x^3 + 2x$

Now, add these expanded terms:

$\frac{dy}{dx} = (6x^4 - 2x^3 + 30x^2 - 10x) + (4x^4 - 2x^3 + 2x)$

Combine like terms:

$\frac{dy}{dx} = (6x^4 + 4x^4) + (-2x^3 - 2x^3) + 30x^2 + (-10x + 2x)$

$\frac{dy}{dx} = 10x^4 - 4x^3 + 30x^2 - 8x$

Part 2: Verify the result by expanding first and then differentiating.

Expand the original product: $y = (x^2 + 5)(2x^3 - x^2 + 1)$

$y = x^2(2x^3 - x^2 + 1) + 5(2x^3 - x^2 + 1)$

$y = (2x^5 - x^4 + x^2) + (10x^3 - 5x^2 + 5)$

$y = 2x^5 - x^4 + 10x^3 - 4x^2 + 5$

Now, differentiate this expanded polynomial term by term:

$\frac{dy}{dx} = \frac{d}{dx}(2x^5 - x^4 + 10x^3 - 4x^2 + 5)$

$\frac{dy}{dx} = 2(5x^4) - 4x^3 + 10(3x^2) - 4(2x) + 0$

$\frac{dy}{dx} = 10x^4 - 4x^3 + 30x^2 - 8x$

The result from both methods matches, confirming the derivative is $10x^4 - 4x^3 + 30x^2 - 8x$.

The given function is $f(x) = |x-2| + 3$.

Domain:

The absolute value function $|x-2|$ is defined for all real numbers $x$. Adding a constant (3) does not introduce any restrictions on the input $x$. Therefore, the domain of $f(x)$ is all real numbers.

Domain: $\{x \in \mathbb{R}\}$

In interval notation: $(-\infty, \infty)$

Range:

The absolute value function $|x-2|$ always produces a non-negative output, meaning $|x-2| \geq 0$ for all real numbers $x$.

When we add 3 to this non-negative value, the smallest possible output is $0 + 3 = 3$.

Therefore, $f(x) = |x-2| + 3 \geq 3$.

The range is all real numbers greater than or equal to 3.

Range: $\{y \in \mathbb{R} \mid y \geq 3\}$

In interval notation: $[3, \infty)$

Sketching the graph:

The graph of $f(x) = |x-2| + 3$ is a transformation of the basic absolute value function $y = |x|$.

• The term $|x-2|$ shifts the graph of $y=|x|$ two units to the right. The vertex moves from (0,0) to (2,0).
• The term $+3$ shifts the graph vertically 3 units up. The vertex moves from (2,0) to (2,3).

The graph is V-shaped with its vertex at $(2, 3)$.

• For $x \geq 2$, $x-2 \geq 0$, so $|x-2| = x-2$. Thus, $f(x) = (x-2) + 3 = x+1$. This is a line with a slope of 1 for $x \geq 2$.
• For $x < 2$, $x-2 < 0$, so $|x-2| = -(x-2) = -x+2$. Thus, $f(x) = (-x+2) + 3 = -x+5$. This is a line with a slope of -1 for $x < 2$.

The graph will have its lowest point (vertex) at $(2, 3)$.

*(A visual representation would show a V-shaped graph with its vertex at (2, 3), opening upwards.)*

We need to evaluate the limit: $\lim\limits_{x \to a} \frac{\sqrt{x} - \sqrt{a}}{x - a}$.

If we substitute $x=a$ directly, we get $\frac{\sqrt{a} - \sqrt{a}}{a - a} = \frac{0}{0}$, which is an indeterminate form.

To evaluate this limit, we can rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the numerator, which is $\sqrt{x} + \sqrt{a}$.

Multiply the expression by $\frac{\sqrt{x} + \sqrt{a}}{\sqrt{x} + \sqrt{a}}$:

$\lim\limits_{x \to a} \frac{\sqrt{x} - \sqrt{a}}{x - a} \times \frac{\sqrt{x} + \sqrt{a}}{\sqrt{x} + \sqrt{a}}$

Multiply the numerators:

$(\sqrt{x} - \sqrt{a})(\sqrt{x} + \sqrt{a}) = (\sqrt{x})^2 - (\sqrt{a})^2 = x - a$

Multiply the denominators:

$(x - a)(\sqrt{x} + \sqrt{a})$

So, the expression becomes:

$\lim\limits_{x \to a} \frac{x - a}{(x - a)(\sqrt{x} + \sqrt{a})}$

Since $x \to a$, $x$ is not equal to $a$, so $(x - a)$ is not zero. We can cancel out the $(x - a)$ terms:

$\lim\limits_{x \to a} \frac{1}{\sqrt{x} + \sqrt{a}}$

Now, substitute $x = a$ into the simplified expression:

$\frac{1}{\sqrt{a} + \sqrt{a}} = \frac{1}{2\sqrt{a}}$

Alternatively, we can recognize this limit as the definition of the derivative of $f(x) = \sqrt{x}$ at the point $x=a$.

The derivative of $f(x) = \sqrt{x} = x^{1/2}$ is $f'(x) = \frac{1}{2}x^{1/2 - 1} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$.

Evaluating this at $x=a$ gives $f'(a) = \frac{1}{2\sqrt{a}}$.

Therefore, the value of the limit is $\frac{1}{2\sqrt{a}}$.

Part 1: Find the derivative of $f(x) = \sqrt{x}$ using the definition.

The definition of the derivative is $f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$.

Given $f(x) = \sqrt{x}$.

First, find $f(x+h)$: $f(x+h) = \sqrt{x+h}$.

Substitute $f(x+h)$ and $f(x)$ into the definition:

$f'(x) = \lim\limits_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$

To simplify this expression, we rationalize the numerator by multiplying by its conjugate, $\sqrt{x+h} + \sqrt{x}$:

$f'(x) = \lim\limits_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \times \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}$

Multiply the numerators using $(a-b)(a+b) = a^2 - b^2$:

$(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x}) = (\sqrt{x+h})^2 - (\sqrt{x})^2 = (x+h) - x = h$

Multiply the denominators:

$h(\sqrt{x+h} + \sqrt{x})$

So, the expression becomes:

$f'(x) = \lim\limits_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}$

Since $h \to 0$, $h \neq 0$, we can cancel out the $h$ terms:

$f'(x) = \lim\limits_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}$

Now, substitute $h=0$ into the simplified expression:

$f'(x) = \frac{1}{\sqrt{x+0} + \sqrt{x}} = \frac{1}{\sqrt{x} + \sqrt{x}} = \frac{1}{2\sqrt{x}}$

So, the derivative of $f(x) = \sqrt{x}$ is $f'(x) = \frac{1}{2\sqrt{x}}$.

Part 2: Find the slope of the tangent line to the curve $y = \sqrt{x}$ at $x=4$.

The slope of the tangent line at a point is given by the value of the derivative at that point.

We found the derivative to be $f'(x) = \frac{1}{2\sqrt{x}}$.

To find the slope at $x=4$, substitute $x=4$ into the derivative:

Slope $m = f'(4) = \frac{1}{2\sqrt{4}}$

$m = \frac{1}{2(2)}$

$m = \frac{1}{4}$

The slope of the tangent line to the curve $y = \sqrt{x}$ at $x=4$ is $\frac{1}{4}$.

For a function $f(x)$ to be continuous at a point $x=c$, three conditions must be met:

1. $f(c)$ must be defined.
2. $\lim\limits_{x \to c} f(x)$ must exist.
3. $\lim\limits_{x \to c} f(x) = f(c)$.

In this problem, we check for continuity at $c=3$. The function is defined piecewise:

$f(x) = \begin{cases} \frac{x^2 - 9}{x - 3} & , & x \neq 3 \\ 6 & , & x = 3 \end{cases}$

1. Check if $f(3)$ is defined:

The definition for $x=3$ is explicitly given as $f(3) = 6$.

So, $f(3)$ is defined and $f(3) = 6$.

2. Check if $\lim\limits_{x \to 3} f(x)$ exists:

To find the limit as $x$ approaches 3, we use the part of the function where $x \neq 3$:

$\lim\limits_{x \to 3} f(x) = \lim\limits_{x \to 3} \frac{x^2 - 9}{x - 3}$

If we substitute $x=3$, we get $\frac{3^2 - 9}{3 - 3} = \frac{9-9}{0} = \frac{0}{0}$, which is an indeterminate form.

We can simplify the expression by factoring the numerator ($x^2 - 9 = (x-3)(x+3)$):

$\lim\limits_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3}$

Since $x \to 3$, $x \neq 3$, so we can cancel out the $(x-3)$ terms:

$\lim\limits_{x \to 3} (x + 3)$

Now, substitute $x=3$: $3 + 3 = 6$.

Thus, the limit exists and $\lim\limits_{x \to 3} f(x) = 6$.

3. Check if $\lim\limits_{x \to 3} f(x) = f(3)$:

From step 1, $f(3) = 6$.

From step 2, $\lim\limits_{x \to 3} f(x) = 6$.

Since $\lim\limits_{x \to 3} f(x) = f(3)$, the third condition for continuity is satisfied.

Conclusion:

All three conditions for continuity are met. Therefore, the function $f(x)$ is continuous at $x=3$.

To find the derivative of $y = (\frac{x+1}{x-1})^3$, we will use the chain rule in conjunction with the quotient rule.

The chain rule states that if $y = [g(x)]^n$, then $\frac{dy}{dx} = n[g(x)]^{n-1} \cdot g'(x)$.

In this case, $g(x) = \frac{x+1}{x-1}$ and $n=3$.

First, let's find the derivative of $g(x) = \frac{x+1}{x-1}$ using the quotient rule. The quotient rule is $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$.

Let $u = x+1$ and $v = x-1$.

$\frac{du}{dx} = \frac{d}{dx}(x+1) = 1$

$\frac{dv}{dx} = \frac{d}{dx}(x-1) = 1$

Applying the quotient rule to $g(x)$:

$g'(x) = \frac{(x-1)(1) - (x+1)(1)}{(x-1)^2}$

$g'(x) = \frac{x - 1 - (x + 1)}{(x-1)^2}$

$g'(x) = \frac{x - 1 - x - 1}{(x-1)^2}$

$g'(x) = \frac{-2}{(x-1)^2}$

Now, apply the chain rule to $y = [g(x)]^3$:

$\frac{dy}{dx} = 3[g(x)]^{3-1} \cdot g'(x)$

$\frac{dy}{dx} = 3\left(\frac{x+1}{x-1}\right)^2 \cdot \left(\frac{-2}{(x-1)^2}\right)$

Now, simplify the expression:

$\frac{dy}{dx} = 3 \cdot \frac{(x+1)^2}{(x-1)^2} \cdot \frac{-2}{(x-1)^2}$

$\frac{dy}{dx} = \frac{3 \cdot (x+1)^2 \cdot (-2)}{(x-1)^2 \cdot (x-1)^2}$

$\frac{dy}{dx} = \frac{-6 (x+1)^2}{(x-1)^4}$

Thus, the derivative of $y = (\frac{x+1}{x-1})^3$ is $\frac{-6 (x+1)^2}{(x-1)^4}$.

The given function is $f(x) = \frac{1}{\sqrt{x^2 - 4}}$.

Domain:

For the function to be defined, the expression under the square root must be strictly positive (greater than zero), because it is in the denominator.

So, we must have: $x^2 - 4 > 0$.

To solve this inequality, first find the values of $x$ for which $x^2 - 4 = 0$.

$x^2 - 4 = 0$

This is a difference of squares: $(x-2)(x+2) = 0$.

The solutions are $x=2$ and $x=-2$.

These values divide the number line into three intervals: $(-\infty, -2)$, $(-2, 2)$, and $(2, \infty)$. We test a value from each interval to see where $x^2 - 4 > 0$.

• For $(-\infty, -2)$, let $x=-3$. Then $(-3)^2 - 4 = 9 - 4 = 5$. Since $5 > 0$, this interval is part of the domain.
• For $(-2, 2)$, let $x=0$. Then $(0)^2 - 4 = 0 - 4 = -4$. Since $-4 \ngtr 0$, this interval is not part of the domain.
• For $(2, \infty)$, let $x=3$. Then $(3)^2 - 4 = 9 - 4 = 5$. Since $5 > 0$, this interval is part of the domain.

Therefore, the domain is $x < -2$ or $x > 2$.

Domain: $\{x \in \mathbb{R} \mid x < -2 \text{ or } x > 2\}$

In interval notation: $(-\infty, -2) \cup (2, \infty)$

Range:

The term $\sqrt{x^2 - 4}$ will always be positive because $x^2 - 4 > 0$ and we are taking the principal square root.

As $x$ approaches $\infty$ or $-\infty$, $x^2 - 4$ approaches $\infty$, so $\sqrt{x^2 - 4}$ approaches $\infty$. This means $\frac{1}{\sqrt{x^2 - 4}}$ approaches $0$ from the positive side.

As $x$ approaches $2$ from the right ($x \to 2^+$), $x^2 - 4$ approaches $0$ from the positive side, so $\sqrt{x^2 - 4}$ approaches $0$ from the positive side. This means $\frac{1}{\sqrt{x^2 - 4}}$ approaches $\infty$.

As $x$ approaches $-2$ from the left ($x \to -2^-$), $x^2 - 4$ approaches $0$ from the positive side, so $\sqrt{x^2 - 4}$ approaches $0$ from the positive side. This means $\frac{1}{\sqrt{x^2 - 4}}$ approaches $\infty$.

The function $f(x)$ can take any positive value. It approaches 0 but never reaches it.

Range: $\{y \in \mathbb{R} \mid y > 0\}$

In interval notation: $(0, \infty)$

We need to evaluate the limit: $\lim\limits_{x \to 0} \frac{\sqrt{a+x} - \sqrt{a-x}}{x}$.

If we substitute $x=0$ directly, we get $\frac{\sqrt{a+0} - \sqrt{a-0}}{0} = \frac{\sqrt{a} - \sqrt{a}}{0} = \frac{0}{0}$, which is an indeterminate form.

To evaluate this limit, we can rationalize the numerator by multiplying the numerator and the denominator by the conjugate of the numerator, which is $\sqrt{a+x} + \sqrt{a-x}$.

Multiply the expression by $\frac{\sqrt{a+x} + \sqrt{a-x}}{\sqrt{a+x} + \sqrt{a-x}}$:

$\lim\limits_{x \to 0} \frac{\sqrt{a+x} - \sqrt{a-x}}{x} \times \frac{\sqrt{a+x} + \sqrt{a-x}}{\sqrt{a+x} + \sqrt{a-x}}$

Multiply the numerators using the difference of squares formula $(A-B)(A+B) = A^2 - B^2$:

$(\sqrt{a+x} - \sqrt{a-x})(\sqrt{a+x} + \sqrt{a-x}) = (\sqrt{a+x})^2 - (\sqrt{a-x})^2$

= $(a+x) - (a-x)$

= $a+x - a+x$

= $2x$

Multiply the denominators:

$x(\sqrt{a+x} + \sqrt{a-x})$

So, the expression becomes:

$\lim\limits_{x \to 0} \frac{2x}{x(\sqrt{a+x} + \sqrt{a-x})}$

Since $x \to 0$, $x \neq 0$, we can cancel out the $x$ terms:

$\lim\limits_{x \to 0} \frac{2}{\sqrt{a+x} + \sqrt{a-x}}$

Now, substitute $x = 0$ into the simplified expression:

$\frac{2}{\sqrt{a+0} + \sqrt{a-0}} = \frac{2}{\sqrt{a} + \sqrt{a}}$

= $\frac{2}{2\sqrt{a}}$

= $\frac{1}{\sqrt{a}}$

Therefore, the value of the limit is $\frac{1}{\sqrt{a}}$. Note that this requires $a > 0$ for $\sqrt{a}$ to be defined and non-zero.

Part 1: Using first principles, find the derivative of $f(x) = x^3$.

The definition of the derivative (first principles) is:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Given $f(x) = x^3$.

First, find $f(x+h)$:

$f(x+h) = (x+h)^3$

Expand $(x+h)^3$ using the binomial expansion formula $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$:

$f(x+h) = x^3 + 3x^2h + 3xh^2 + h^3$

Substitute $f(x+h)$ and $f(x)$ into the definition of the derivative:

$f'(x) = \lim\limits_{h \to 0} \frac{(x^3 + 3x^2h + 3xh^2 + h^3) - x^3}{h}$

Simplify the numerator:

$f'(x) = \lim\limits_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h}$

Factor out $h$ from the numerator:

$f'(x) = \lim\limits_{h \to 0} \frac{h(3x^2 + 3xh + h^2)}{h}$

Since $h \to 0$, $h \neq 0$, we can cancel the $h$ terms:

$f'(x) = \lim\limits_{h \to 0} (3x^2 + 3xh + h^2)$

Now, substitute $h=0$ into the simplified expression:

$f'(x) = 3x^2 + 3x(0) + (0)^2 = 3x^2 + 0 + 0 = 3x^2$

So, the derivative of $f(x) = x^3$ is $f'(x) = 3x^2$.

Part 2: Find the slope of the tangent to the curve $y = x^3$ at $x=-1$.

The slope of the tangent line at a point is given by the value of the derivative at that point.

We found the derivative to be $f'(x) = 3x^2$.

To find the slope at $x=-1$, substitute $x=-1$ into the derivative:

Slope $m = f'(-1) = 3(-1)^2 = 3(1) = 3$

The slope of the tangent line to the curve $y = x^3$ at $x=-1$ is 3.

For a function $f(x)$ to be continuous at a point $x=c$, the following three conditions must be met:

1. $f(c)$ must be defined.
2. $\lim\limits_{x \to c} f(x)$ must exist.
3. $\lim\limits_{x \to c} f(x) = f(c)$.

In this problem, we check for continuity at $c=1$. The function is defined piecewise:

$f(x) = \begin{cases} ax+b & , & x < 1 \\ 5 & , & x = 1 \\ 2x-3 & , & x > 1 \end{cases}$

1. Check if $f(1)$ is defined:

The definition for $x=1$ is explicitly given as $f(1) = 5$.

So, $f(1)$ is defined and $f(1) = 5$.

2. Check if $\lim\limits_{x \to 1} f(x)$ exists:

For the limit to exist, the left-hand limit and the right-hand limit must be equal.

Left-hand limit: As $x$ approaches 1 from the left ($x < 1$), we use $f(x) = ax+b$.

$\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^-} (ax+b) = a(1) + b = a+b$.

Right-hand limit: As $x$ approaches 1 from the right ($x > 1$), we use $f(x) = 2x-3$.

$\lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1^+} (2x-3) = 2(1)-3 = 2-3 = -1$.

For the limit to exist, the left-hand limit must equal the right-hand limit:

3. Check if $\lim\limits_{x \to 1} f(x) = f(1)$:

From step 1, $f(1) = 5$.

From step 2, for the limit to exist, it must be equal to $-1$ (from the right-hand limit). So, $\lim\limits_{x \to 1} f(x) = -1$.

For continuity, we must have $\lim\limits_{x \to 1} f(x) = f(1)$.

This means $-1 = 5$.

However, we have a contradiction: $-1 \neq 5$. This implies that there are no values of $a$ and $b$ that can make this function continuous at $x=1$ as defined, because the value of the function at $x=1$ (which is 5) does not match the limit as $x$ approaches 1 (which must be -1 for the left and right limits to be equal).

Let's re-examine the conditions. For continuity, we need:

Left-hand limit = Right-hand limit = $f(1)$

From the definition:

$\lim\limits_{x \to 1^-} f(x) = a+b$

$\lim\limits_{x \to 1^+} f(x) = -1$

$f(1) = 5$

For continuity, we require:

$a+b = -1$ (for the left and right limits to be equal)

AND

The limit must equal $f(1)$. So, $-1 = 5$.

Since $-1 \neq 5$, there is a fundamental inconsistency in the problem statement as it prevents continuity. If the function were to be continuous, the value of $f(1)$ must be equal to the limit. The limit requires $a+b = -1$, but $f(1)$ is fixed at 5.

There are no values of $a$ and $b$ that can make this function continuous at $x=1$.

First, rewrite the function using exponent notation:

$y = \left(x^2 + x^{-2}\right)^{1/2}$

To find the derivative of this function, we will use the chain rule. The chain rule states that if $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$.

Let $u = x^2 + x^{-2}$ and $y = u^{1/2}$.

Find the derivative of $y$ with respect to $u$ ($f'(u)$):

$\frac{dy}{du} = \frac{d}{du}(u^{1/2})$

Using the power rule, $\frac{d}{du}(u^n) = nu^{n-1}$:

$\frac{dy}{du} = \frac{1}{2}u^{\frac{1}{2}-1} = \frac{1}{2}u^{-\frac{1}{2}}$

Find the derivative of $u$ with respect to $x$ ($g'(x)$):

$\frac{du}{dx} = \frac{d}{dx}(x^2 + x^{-2})$

Using the power rule:

$\frac{du}{dx} = 2x^{2-1} + (-2)x^{-2-1} = 2x - 2x^{-3}$

Now, apply the chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

$\frac{dy}{dx} = \left(\frac{1}{2}u^{-\frac{1}{2}}\right) \cdot (2x - 2x^{-3})$

Substitute back $u = x^2 + x^{-2}$:

$\frac{dy}{dx} = \frac{1}{2}(x^2 + x^{-2})^{-\frac{1}{2}} (2x - 2x^{-3})$

Factor out a 2 from the second term:

$\frac{dy}{dx} = \frac{1}{2}(x^2 + x^{-2})^{-\frac{1}{2}} \cdot 2(x - x^{-3})$

$\frac{dy}{dx} = (x^2 + x^{-2})^{-\frac{1}{2}} (x - x^{-3})$

Rewrite with positive exponents and in fractional form:

$\frac{dy}{dx} = \frac{x - \frac{1}{x^3}}{\sqrt{x^2 + \frac{1}{x^2}}}$

To simplify the numerator, find a common denominator:

$x - \frac{1}{x^3} = \frac{x \cdot x^3}{x^3} - \frac{1}{x^3} = \frac{x^4 - 1}{x^3}$

Rewrite the denominator with a common denominator:

$\sqrt{x^2 + \frac{1}{x^2}} = \sqrt{\frac{x^4 + 1}{x^2}} = \frac{\sqrt{x^4 + 1}}{\sqrt{x^2}} = \frac{\sqrt{x^4 + 1}}{|x|}$

Substitute these back into the derivative expression:

$\frac{dy}{dx} = \frac{\frac{x^4 - 1}{x^3}}{\frac{\sqrt{x^4 + 1}}{|x|}}$

$\frac{dy}{dx} = \frac{x^4 - 1}{x^3} \cdot \frac{|x|}{\sqrt{x^4 + 1}}$

If we assume $x>0$, then $|x|=x$:

$\frac{dy}{dx} = \frac{x^4 - 1}{x^3} \cdot \frac{x}{\sqrt{x^4 + 1}} = \frac{x(x^4 - 1)}{x^3\sqrt{x^4 + 1}} = \frac{x^4 - 1}{x^2\sqrt{x^4 + 1}}$

Let's go back to the expression $\frac{dy}{dx} = (x^2 + x^{-2})^{-\frac{1}{2}} (x - x^{-3})$ and simplify it differently.

$\frac{dy}{dx} = \frac{1}{\sqrt{x^2 + \frac{1}{x^2}}} \cdot \left(x - \frac{1}{x^3}\right)$

$\frac{dy}{dx} = \frac{1}{\sqrt{\frac{x^4+1}{x^2}}} \cdot \left(\frac{x^4-1}{x^3}\right)$

$\frac{dy}{dx} = \frac{|x|}{\sqrt{x^4+1}} \cdot \frac{x^4-1}{x^3}$

Assuming $x>0$, $|x|=x$:

$\frac{dy}{dx} = \frac{x}{\sqrt{x^4+1}} \cdot \frac{x^4-1}{x^3} = \frac{x(x^4-1)}{x^3\sqrt{x^4+1}} = \frac{x^4-1}{x^2\sqrt{x^4+1}}$

Thus, the derivative is $\frac{x^4 - 1}{x^2\sqrt{x^4 + 1}}$ (for $x>0$).

Let $V$ be the volume of the sphere, $A$ be its surface area, and $r$ be its radius. We are given the following information:

• The rate of increase of the volume with respect to time ($t$) is $\frac{dV}{dt} = 4\pi$ cm$^3$/sec.
• We need to find the rate of increase of the surface area, $\frac{dA}{dt}$, when $r = 2$ cm.

The formulas for the volume and surface area of a sphere are:

Volume: $V = \frac{4}{3}\pi r^3$

Surface Area: $A = 4\pi r^2$

Step 1: Find the relationship between $\frac{dV}{dt}$ and $\frac{dr}{dt}$.

Differentiate the volume formula with respect to time $t$ using the chain rule:

$\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)$

$\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2 \cdot \frac{dr}{dt}$

$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$

We are given $\frac{dV}{dt} = 4\pi$. Substitute this into the equation:

$4\pi = 4\pi r^2 \frac{dr}{dt}$

Now, we can solve for $\frac{dr}{dt}$. We need the value of $\frac{dr}{dt}$ when $r=2$ cm.

Substitute $r=2$ into the equation:

$4\pi = 4\pi (2)^2 \frac{dr}{dt}$

$4\pi = 4\pi (4) \frac{dr}{dt}$

$4\pi = 16\pi \frac{dr}{dt}$

Solve for $\frac{dr}{dt}$:

$\frac{dr}{dt} = \frac{4\pi}{16\pi} = \frac{1}{4}$ cm/sec.

Step 2: Find the relationship between $\frac{dA}{dt}$ and $\frac{dr}{dt}$.

Differentiate the surface area formula with respect to time $t$ using the chain rule:

$\frac{dA}{dt} = \frac{d}{dt}(4\pi r^2)$

$\frac{dA}{dt} = 4\pi \cdot 2r \cdot \frac{dr}{dt}$

$\frac{dA}{dt} = 8\pi r \frac{dr}{dt}$

Step 3: Calculate $\frac{dA}{dt}$ when $r=2$ cm.

We have $r=2$ cm and we found $\frac{dr}{dt} = \frac{1}{4}$ cm/sec.

Substitute these values into the equation for $\frac{dA}{dt}$:

$\frac{dA}{dt} = 8\pi (2) \left(\frac{1}{4}\right)$

$\frac{dA}{dt} = 16\pi \left(\frac{1}{4}\right)$

$\frac{dA}{dt} = 4\pi$ cm$^2$/sec.

Therefore, the rate at which the surface area is increasing when the radius is 2 cm is $4\pi$ cm$^2$/sec.

The given function is $f(x) = \frac{|x|}{x}$.

Domain:

The domain of a function is the set of all possible input values ($x$-values) for which the function is defined. For $f(x) = \frac{|x|}{x}$, the function is undefined when the denominator is zero.

Setting the denominator to zero:

$x = 0$

Therefore, $x=0$ must be excluded from the domain.

Domain: $\{x \in \mathbb{R} \mid x \neq 0\}$

In interval notation: $(-\infty, 0) \cup (0, \infty)$

Range:

We can analyze the function based on the definition of the absolute value:

• If $x > 0$, then $|x| = x$. So, $f(x) = \frac{x}{x} = 1$.
• If $x < 0$, then $|x| = -x$. So, $f(x) = \frac{-x}{x} = -1$.

The function can only take two values: 1 (when $x>0$) and -1 (when $x<0$).

Range: $\{-1, 1\}$

Sketching the graph:

The graph of $f(x) = \frac{|x|}{x}$ will consist of two horizontal rays:

• For $x > 0$, the graph is the horizontal line $y = 1$. This ray starts at $(0,1)$ but the point $(0,1)$ is not included (open circle).
• For $x < 0$, the graph is the horizontal line $y = -1$. This ray ends at $(0,-1)$ but the point $(0,-1)$ is not included (open circle).

*(A visual representation would show two horizontal lines: one at $y=1$ for $x>0$ and another at $y=-1$ for $x<0$, with open circles at $(0,1)$ and $(0,-1)$.)*

Continuity at $x=0$:

To check for continuity at $x=0$, we apply the three conditions:

1. Is $f(0)$ defined? No, because $x=0$ is not in the domain of the function.

Since the first condition is not met, the function $f(x) = \frac{|x|}{x}$ is discontinuous at $x=0$. This is a non-removable discontinuity (specifically, an infinite discontinuity if we consider the behavior around $x=0$, though more precisely it's a jump discontinuity if we consider the graph piecewise defined, but the domain exclusion makes it simply discontinuous).

To further describe the discontinuity: The left-hand limit as $x \to 0^-$ is $-1$, and the right-hand limit as $x \to 0^+$ is $1$. Since these limits are not equal, the limit at $x=0$ does not exist, which is another reason for the discontinuity.

We need to evaluate the limit: $\lim\limits_{x \to 1} \frac{x^{10} - 1}{x^6 - 1}$.

If we substitute $x=1$ directly, we get $\frac{1^{10} - 1}{1^6 - 1} = \frac{1 - 1}{1 - 1} = \frac{0}{0}$, which is an indeterminate form.

We can solve this using L'Hôpital's Rule or by factoring.

Method 1: Using L'Hôpital's Rule

L'Hôpital's Rule states that if $\lim\limits_{x \to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim\limits_{x \to c} \frac{f(x)}{g(x)} = \lim\limits_{x \to c} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.

Here, $f(x) = x^{10} - 1$ and $g(x) = x^6 - 1$.

Find the derivatives:

$f'(x) = \frac{d}{dx}(x^{10} - 1) = 10x^9$

$g'(x) = \frac{d}{dx}(x^6 - 1) = 6x^5$

Apply L'Hôpital's Rule:

$\lim\limits_{x \to 1} \frac{x^{10} - 1}{x^6 - 1} = \lim\limits_{x \to 1} \frac{10x^9}{6x^5}$

Now, substitute $x=1$ into the simplified expression:

$\frac{10(1)^9}{6(1)^5} = \frac{10}{6} = \frac{5}{3}$

Method 2: Using the definition of the derivative

We can rewrite the limit as:

$\lim\limits_{x \to 1} \frac{x^{10} - 1}{x^6 - 1} = \lim\limits_{x \to 1} \frac{\frac{x^{10} - 1}{x-1}}{\frac{x^6 - 1}{x-1}}$

Recall the definition of the derivative of $f(x) = x^n$ at $x=1$: $f'(1) = \lim\limits_{x \to 1} \frac{x^n - 1^n}{x-1}$.

The derivative of $x^{10}$ is $10x^9$. At $x=1$, this is $10(1)^9 = 10$. So, $\lim\limits_{x \to 1} \frac{x^{10} - 1}{x-1} = 10$.

The derivative of $x^6$ is $6x^5$. At $x=1$, this is $6(1)^5 = 6$. So, $\lim\limits_{x \to 1} \frac{x^6 - 1}{x-1} = 6$.

Now, applying the limit properties:

$\lim\limits_{x \to 1} \frac{\frac{x^{10} - 1}{x-1}}{\frac{x^6 - 1}{x-1}} = \frac{\lim\limits_{x \to 1} \frac{x^{10} - 1}{x-1}}{\lim\limits_{x \to 1} \frac{x^6 - 1}{x-1}} = \frac{10}{6} = \frac{5}{3}$

Therefore, the value of the limit is $\frac{5}{3}$.

The definition of the derivative (first principles) is:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Given $f(x) = \frac{1}{\sqrt{x}}$.

First, find $f(x+h)$: $f(x+h) = \frac{1}{\sqrt{x+h}}$.

Substitute $f(x+h)$ and $f(x)$ into the definition:

$f'(x) = \lim\limits_{h \to 0} \frac{\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}}{h}$

To simplify the numerator, find a common denominator:

$f'(x) = \lim\limits_{h \to 0} \frac{\frac{\sqrt{x} - \sqrt{x+h}}{\sqrt{x}\sqrt{x+h}}}{h}$

$f'(x) = \lim\limits_{h \to 0} \frac{\sqrt{x} - \sqrt{x+h}}{h\sqrt{x}\sqrt{x+h}}$

To evaluate this limit, we rationalize the numerator by multiplying by its conjugate, $\sqrt{x} + \sqrt{x+h}$:

$f'(x) = \lim\limits_{h \to 0} \frac{\sqrt{x} - \sqrt{x+h}}{h\sqrt{x}\sqrt{x+h}} \times \frac{\sqrt{x} + \sqrt{x+h}}{\sqrt{x} + \sqrt{x+h}}$

Multiply the numerators using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$:

$(\sqrt{x} - \sqrt{x+h})(\sqrt{x} + \sqrt{x+h}) = (\sqrt{x})^2 - (\sqrt{x+h})^2 = x - (x+h) = x - x - h = -h$

So the expression becomes:

$f'(x) = \lim\limits_{h \to 0} \frac{-h}{h\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})}$

Since $h \to 0$, $h \neq 0$, we can cancel the $h$ terms:

$f'(x) = \lim\limits_{h \to 0} \frac{-1}{\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})}$

Now, substitute $h=0$ into the simplified expression:

$f'(x) = \frac{-1}{\sqrt{x}\sqrt{x+0}(\sqrt{x} + \sqrt{x+0})}$

$f'(x) = \frac{-1}{\sqrt{x}\sqrt{x}(\sqrt{x} + \sqrt{x})}$

$f'(x) = \frac{-1}{x(2\sqrt{x})}$

$f'(x) = \frac{-1}{2x\sqrt{x}}$

This can also be written using exponents:

$f'(x) = \frac{-1}{2x^{3/2}}$ or $-\frac{1}{2}x^{-3/2}$.

Thus, the derivative of $f(x) = \frac{1}{\sqrt{x}}$ using first principles is $f'(x) = \frac{-1}{2x\sqrt{x}}$.

For a function $f(x)$ to be continuous at a point $x=c$, three conditions must be met:

1. $f(c)$ must be defined.
2. $\lim\limits_{x \to c} f(x)$ must exist.
3. $\lim\limits_{x \to c} f(x) = f(c)$.

In this problem, the function is $f(x) = |x-3|$, and we check for continuity at $c=3$. This is an absolute value function.

1. Check if $f(3)$ is defined:

$f(3) = |3-3| = |0| = 0$.

So, $f(3)$ is defined and $f(3) = 0$.

2. Check if $\lim\limits_{x \to 3} f(x)$ exists:

We need to evaluate the left-hand limit and the right-hand limit.

Left-hand limit: As $x$ approaches 3 from the left ($x < 3$), $x-3$ is negative, so $|x-3| = -(x-3) = 3-x$.

$\lim\limits_{x \to 3^-} f(x) = \lim\limits_{x \to 3^-} |x-3| = \lim\limits_{x \to 3^-} (3-x) = 3 - 3 = 0$.

Right-hand limit: As $x$ approaches 3 from the right ($x > 3$), $x-3$ is positive, so $|x-3| = x-3$.

$\lim\limits_{x \to 3^+} f(x) = \lim\limits_{x \to 3^+} |x-3| = \lim\limits_{x \to 3^+} (x-3) = 3 - 3 = 0$.

Since the left-hand limit and the right-hand limit are equal ($0 = 0$), the limit $\lim\limits_{x \to 3} f(x)$ exists and is equal to 0.

3. Check if $\lim\limits_{x \to 3} f(x) = f(3)$:

From step 1, $f(3) = 0$.

From step 2, $\lim\limits_{x \to 3} f(x) = 0$.

Since $\lim\limits_{x \to 3} f(x) = f(3)$, the third condition for continuity is satisfied.

Conclusion:

All three conditions for continuity are met. Therefore, the function $f(x) = |x-3|$ is continuous at $x=3$.

To find the derivative of $y = \frac{(x^2+3)^4}{(2x-1)^3}$, we will use the quotient rule and the chain rule.

The quotient rule states that if $y = \frac{u}{v}$, then $\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$.

Let $u = (x^2+3)^4$ and $v = (2x-1)^3$.

First, find the derivative of $u$ with respect to $x$, using the chain rule:

Let $w = x^2+3$. Then $u = w^4$.

$\frac{du}{dx} = \frac{du}{dw} \cdot \frac{dw}{dx}$

$\frac{du}{dw} = 4w^3$

$\frac{dw}{dx} = 2x$

So, $\frac{du}{dx} = 4(x^2+3)^3 \cdot (2x) = 8x(x^2+3)^3$.

Next, find the derivative of $v$ with respect to $x$, using the chain rule:

Let $z = 2x-1$. Then $v = z^3$.

$\frac{dv}{dx} = \frac{dv}{dz} \cdot \frac{dz}{dx}$

$\frac{dv}{dz} = 3z^2$

$\frac{dz}{dx} = 2$

So, $\frac{dv}{dx} = 3(2x-1)^2 \cdot (2) = 6(2x-1)^2$.

Now, apply the quotient rule: $\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$

$\frac{dy}{dx} = \frac{(2x-1)^3 [8x(x^2+3)^3] - (x^2+3)^4 [6(2x-1)^2]}{((2x-1)^3)^2}$

Simplify the denominator:

$((2x-1)^3)^2 = (2x-1)^6$

Simplify the numerator by factoring out common terms. The common terms are $(2x-1)^2$ and $(x^2+3)^3$.

Numerator $= (2x-1)^2 (x^2+3)^3 [8x(x^2+3) - 6(x^2+3)^1]$

Numerator $= (2x-1)^2 (x^2+3)^3 [8x^3 + 24x - 6x^2 - 18]$

So, the derivative is:

$\frac{dy}{dx} = \frac{(2x-1)^2 (x^2+3)^3 (8x^3 - 6x^2 + 24x - 18)}{(2x-1)^6}$

Cancel out $(2x-1)^2$ from the numerator and denominator:

$\frac{dy}{dx} = \frac{(x^2+3)^3 (8x^3 - 6x^2 + 24x - 18)}{(2x-1)^4}$

We can factor out a 2 from the polynomial in the numerator:

$8x^3 - 6x^2 + 24x - 18 = 2(4x^3 - 3x^2 + 12x - 9)$

So, the derivative becomes:

$\frac{dy}{dx} = \frac{2(x^2+3)^3 (4x^3 - 3x^2 + 12x - 9)}{(2x-1)^4}$

The position of the particle is given by $s(t) = t^3 - 6t^2 + 9t + 5$.

Velocity:

The velocity $v(t)$ is the first derivative of the position function $s(t)$ with respect to time $t$.

$v(t) = s'(t) = \frac{d}{dt}(t^3 - 6t^2 + 9t + 5)$

$v(t) = 3t^2 - 6(2t) + 9(1) + 0$

$v(t) = 3t^2 - 12t + 9$ m/sec.

To find the velocity at $t=2$ seconds, substitute $t=2$ into the velocity equation:

$v(2) = 3(2)^2 - 12(2) + 9$

$v(2) = 3(4) - 24 + 9$

$v(2) = 12 - 24 + 9$

$v(2) = -12 + 9 = -3$ m/sec.

The velocity of the particle at $t=2$ seconds is $-3$ m/sec.

Acceleration:

The acceleration $a(t)$ is the first derivative of the velocity function $v(t)$ (or the second derivative of the position function $s(t)$) with respect to time $t$.

$a(t) = v'(t) = \frac{d}{dt}(3t^2 - 12t + 9)$

$a(t) = 3(2t) - 12(1) + 0$

$a(t) = 6t - 12$ m/sec$^2$.

To find the acceleration at $t=2$ seconds, substitute $t=2$ into the acceleration equation:

$a(2) = 6(2) - 12$

$a(2) = 12 - 12 = 0$ m/sec$^2$.

The acceleration of the particle at $t=2$ seconds is $0$ m/sec$^2$.

When is the particle at rest?

The particle is at rest when its velocity is zero ($v(t) = 0$).

Set the velocity equation to zero and solve for $t$:

$v(t) = 3t^2 - 12t + 9 = 0$

Divide the entire equation by 3 to simplify:

$t^2 - 4t + 3 = 0$

Factor the quadratic equation:

We need two numbers that multiply to 3 and add to -4. These numbers are -1 and -3.

$(t - 1)(t - 3) = 0$

Set each factor to zero:

$t - 1 = 0 \implies t = 1$ second.

$t - 3 = 0 \implies t = 3$ seconds.

The particle is at rest at $t = 1$ second and $t = 3$ seconds.

The given function is $f(x) = \frac{x-2}{|x-2|}$.

Domain:

The function is undefined when the denominator is zero. The denominator is $|x-2|$.

Set the denominator to zero:

$|x-2| = 0 \implies x-2 = 0 \implies x = 2$.

Thus, $x=2$ is excluded from the domain.

Domain: $\{x \in \mathbb{R} \mid x \neq 2\}$

In interval notation: $(-\infty, 2) \cup (2, \infty)$

Range:

We analyze the function based on the definition of the absolute value:

• If $x > 2$, then $x-2 > 0$, so $|x-2| = x-2$. $f(x) = \frac{x-2}{x-2} = 1$.
• If $x < 2$, then $x-2 < 0$, so $|x-2| = -(x-2)$. $f(x) = \frac{x-2}{-(x-2)} = -1$.

The function can only take two values: 1 (when $x>2$) and -1 (when $x<2$).

Range: $\{-1, 1\}$

Graph:

The graph of $f(x) = \frac{x-2}{|x-2|}$ consists of two horizontal rays:

• For $x > 2$, the graph is the horizontal line $y = 1$. This ray starts at $(2,1)$ but the point $(2,1)$ is not included (open circle).
• For $x < 2$, the graph is the horizontal line $y = -1$. This ray ends at $(2,-1)$ but the point $(2,-1)$ is not included (open circle).

*(A visual representation would show two horizontal lines: one at $y=1$ for $x>2$ and another at $y=-1$ for $x<2$, with open circles at $(2,1)$ and $(2,-1)$.)*

Continuity:

The function is undefined at $x=2$ (as determined by the domain). Therefore, it cannot be continuous at $x=2$. This is a discontinuity because the function is not defined at $x=2$. It is a type of jump discontinuity (or more accurately, a discontinuity due to an undefined point, with a jump in function value if we were to consider limits). The left-hand limit is -1, and the right-hand limit is 1. Since these limits are not equal, the limit at $x=2$ does not exist, and the function is discontinuous at $x=2$. This is a non-removable discontinuity.

We need to evaluate the limit: $\lim\limits_{x \to 5} \frac{x^3 - 125}{x^2 - 25}$.

If we substitute $x=5$ directly into the expression, we get:

$\frac{5^3 - 125}{5^2 - 25} = \frac{125 - 125}{25 - 25} = \frac{0}{0}$

This is an indeterminate form, so we need to simplify the expression by factoring.

Factor the numerator: $x^3 - 125$ is a difference of cubes ($a^3 - b^3 = (a-b)(a^2+ab+b^2)$) where $a=x$ and $b=5$.

$x^3 - 125 = x^3 - 5^3 = (x - 5)(x^2 + 5x + 5^2) = (x - 5)(x^2 + 5x + 25)$

Factor the denominator: $x^2 - 25$ is a difference of squares ($a^2 - b^2 = (a-b)(a+b)$) where $a=x$ and $b=5$.

$x^2 - 25 = x^2 - 5^2 = (x - 5)(x + 5)$

Now, substitute the factored forms back into the limit expression:

$\lim\limits_{x \to 5} \frac{(x - 5)(x^2 + 5x + 25)}{(x - 5)(x + 5)}$

Since $x \to 5$, $x \neq 5$, so $(x - 5)$ is not zero. We can cancel out the $(x - 5)$ term:

$\lim\limits_{x \to 5} \frac{x^2 + 5x + 25}{x + 5}$

Now, substitute $x = 5$ into the simplified expression:

$\frac{5^2 + 5(5) + 25}{5 + 5}$

= $\frac{25 + 25 + 25}{10}$

= $\frac{75}{10}$

Simplify the fraction:

= $\frac{15}{2}$

Therefore, the value of the limit is $\frac{15}{2}$.

We will solve the revised question: Find the derivative of $y = x^3 (2x+5)^4$.

To find the derivative of this product of two functions, we will use the product rule and the chain rule.

The product rule states that if $y = uv$, then $\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$.

Let $u = x^3$ and $v = (2x+5)^4$.

First, find the derivative of $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(x^3) = 3x^2$

Next, find the derivative of $v$ with respect to $x$. We need the chain rule here:

Let $w = 2x+5$. Then $v = w^4$.

$\frac{dv}{dx} = \frac{dv}{dw} \cdot \frac{dw}{dx}$

$\frac{dv}{dw} = 4w^3$

$\frac{dw}{dx} = \frac{d}{dx}(2x+5) = 2$

So, $\frac{dv}{dx} = 4w^3 \cdot 2 = 8w^3$. Substituting $w = 2x+5$ back:

$\frac{dv}{dx} = 8(2x+5)^3$.

Now, apply the product rule: $\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$

$\frac{dy}{dx} = x^3 \cdot [8(2x+5)^3] + (2x+5)^4 \cdot [3x^2]$

To simplify, we can factor out common terms. The common terms are $x^2$ and $(2x+5)^3$.

$\frac{dy}{dx} = x^2(2x+5)^3 [8x + 3(2x+5)]$

Simplify the expression inside the brackets:

$8x + 3(2x+5) = 8x + 6x + 15 = 14x + 15$

Substitute this back into the derivative:

$\frac{dy}{dx} = x^2(2x+5)^3 (14x + 15)$

Thus, the derivative of $y = x^3 (2x+5)^4$ is $x^2(2x+5)^3 (14x + 15)$.

The definition of the derivative (first principles) is:

$f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Given $f(x) = x^n$, where $n$ is a positive integer.

First, find $f(x+h)$: $f(x+h) = (x+h)^n$.

Substitute $f(x+h)$ and $f(x)$ into the definition:

$f'(x) = \lim\limits_{h \to 0} \frac{(x+h)^n - x^n}{h}$

To expand $(x+h)^n$, we use the Binomial Theorem:

$(x+h)^n = \binom{n}{0}x^n h^0 + \binom{n}{1}x^{n-1} h^1 + \binom{n}{2}x^{n-2} h^2 + \dots + \binom{n}{n-1}x^1 h^{n-1} + \binom{n}{n}x^0 h^n$

$(x+h)^n = x^n + nx^{n-1}h + \frac{n(n-1)}{2!}x^{n-2}h^2 + \dots + nxh^{n-1} + h^n$

Now, substitute this expansion into the derivative formula:

$f'(x) = \lim\limits_{h \to 0} \frac{(x^n + nx^{n-1}h + \frac{n(n-1)}{2!}x^{n-2}h^2 + \dots + nxh^{n-1} + h^n) - x^n}{h}$

Simplify the numerator:

$f'(x) = \lim\limits_{h \to 0} \frac{nx^{n-1}h + \frac{n(n-1)}{2!}x^{n-2}h^2 + \dots + nxh^{n-1} + h^n}{h}$

Factor out $h$ from each term in the numerator:

$f'(x) = \lim\limits_{h \to 0} \frac{h \left( nx^{n-1} + \frac{n(n-1)}{2!}x^{n-2}h + \dots + nxh^{n-2} + h^{n-1} \right)}{h}$

Since $h \to 0$, $h \neq 0$, we can cancel the $h$ terms:

$f'(x) = \lim\limits_{h \to 0} \left( nx^{n-1} + \frac{n(n-1)}{2!}x^{n-2}h + \dots + nxh^{n-2} + h^{n-1} \right)$

Now, substitute $h=0$ into the expression. All terms containing $h$ will become zero:

$f'(x) = nx^{n-1} + 0 + \dots + 0 + 0$

$f'(x) = nx^{n-1}$

Thus, using first principles, the derivative of $f(x) = x^n$ (where $n$ is a positive integer) is $f'(x) = nx^{n-1}$.

We will solve the revised question, addressing both continuity and differentiability.

The function is $f(x) = \begin{cases} ax+3 & , & x \leq 2 \\ x^2 + b & , & x > 2 \end{cases}$.

Part 1: Continuity at $x=2$.

For $f(x)$ to be continuous at $x=2$, we need:

1. $f(2)$ to be defined.
2. $\lim\limits_{x \to 2} f(x)$ to exist.
3. $\lim\limits_{x \to 2} f(x) = f(2)$.

1. $f(2)$ is defined:

Using the first case ($x \leq 2$), $f(2) = a(2) + 3 = 2a + 3$.

2. $\lim\limits_{x \to 2} f(x)$ exists:

This requires the left-hand limit and the right-hand limit to be equal.

Left-hand limit: $\lim\limits_{x \to 2^-} f(x) = \lim\limits_{x \to 2^-} (ax+3) = a(2) + 3 = 2a+3$.

Right-hand limit: $\lim\limits_{x \to 2^+} f(x) = \lim\limits_{x \to 2^+} (x^2+b) = (2)^2 + b = 4+b$.

For continuity, these must be equal:

3. $\lim\limits_{x \to 2} f(x) = f(2)$:

From step 1, $f(2) = 2a+3$.

From step 2, for the limit to exist, it must be equal to $2a+3$ (from the left) and $4+b$ (from the right). For continuity, these must all be equal.

So, we need $2a+3 = 4+b$. This is the same condition as in step 2.

The condition for continuity at $x=2$ is $2a+3 = 4+b$, which can be rewritten as $2a - b = 1$.

There is one equation with two unknowns ($a$ and $b$), meaning there are infinitely many pairs of $(a, b)$ that satisfy continuity. For example, if $a=1$, then $2(1)-b=1 \implies 2-b=1 \implies b=1$. If $a=0$, then $2(0)-b=1 \implies -b=1 \implies b=-1$.

Part 2: Differentiability at $x=2$.

For a function to be differentiable at a point, it must first be continuous at that point. So, $2a-b=1$ must hold.

Additionally, the derivative from the left must equal the derivative from the right at $x=2$.

First, find the derivatives of each piece:

For $x < 2$, $f(x) = ax+3$, so $f'(x) = a$.

For $x > 2$, $f(x) = x^2+b$, so $f'(x) = 2x$.

Now, we need the left-hand derivative and the right-hand derivative at $x=2$ to be equal.

Left-hand derivative at $x=2$: $\lim\limits_{h \to 0^-} \frac{f(2+h) - f(2)}{h}$

Using the derivative of the first piece: $f'(2^-) = a$.

Right-hand derivative at $x=2$: $\lim\limits_{h \to 0^+} \frac{f(2+h) - f(2)}{h}$

Using the derivative of the second piece: $f'(2^+) = 2(2) = 4$.

For differentiability, these must be equal:

Now we have a system of two equations with two unknowns:

1) $2a - b = 1$

2) $a = 4$

Substitute $a=4$ into the first equation:

$2(4) - b = 1$

$8 - b = 1$

$b = 8 - 1$

$b = 7$

So, for the function to be differentiable at $x=2$, we must have $a=4$ and $b=7$. (These values also ensure continuity, as $2(4) - 7 = 8 - 7 = 1$, which satisfies the continuity condition).

Summary:

• For continuity at $x=2$, the condition is $2a - b = 1$.
• For differentiability at $x=2$ (which also implies continuity), the values are $a=4$ and $b=7$.