Chapter 6 Application of Derivatives (Concepts)

Derivatives as a Rate Measure

One of the most fundamental applications of derivatives is to measure the rate at which one quantity changes with respect to another. When we talk about how fast something is changing, we are usually referring to its rate of change. The derivative provides the instantaneous rate of change, which is the rate of change at a very specific point or instant in time or with respect to another variable. This is in contrast to the average rate of change, which is calculated over an interval.

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Instantaneous Rate of Change

If a quantity $y$ is related to another quantity $x$ through a function, say $y = f(x)$, then the derivative $\frac{dy}{dx}$ represents the instantaneous rate of change of $y$ with respect to $x$. This essentially tells us how much $y$ is changing for a tiny change in $x$, at a particular value of $x$.

Recall the formal definition of the derivative of $y = f(x)$ with respect to $x$ from the first principles (limits):

In this definition, $\Delta x$ represents a small (increment) change in the independent variable $x$, and $\Delta y = f(x+\Delta x) - f(x)$ represents the corresponding change in the dependent variable $y$. The ratio $\frac{\Delta y}{\Delta x}$ calculates the average rate of change of $y$ with respect to $x$ over the interval $[x, x+\Delta x]$ (or $[x+\Delta x, x]$ if $\Delta x$ is negative). By taking the limit as $\Delta x \to 0$, we are finding the rate of change of $y$ with respect to $x$ at the exact point $x$, which is the instantaneous rate of change.

Geometrically, the derivative $\frac{dy}{dx}$ evaluated at a specific point $(x_0, y_0)$ on the curve $y = f(x)$ gives the slope of the tangent line to the curve at that point. A larger magnitude of the derivative indicates a faster rate of change (steeper slope), while a smaller magnitude indicates a slower rate of change (gentler slope).

Often, quantities change with respect to time, $t$. If $y$ is a quantity that changes over time, i.e., $y = y(t)$, then $\frac{dy}{dt}$ represents the instantaneous rate of change of $y$ with respect to time $t$. This is a ubiquitous concept in various fields:

• In Physics, velocity $v$ is the rate of change of displacement $s$ with respect to time $t$ ($v = \frac{ds}{dt}$), and acceleration $a$ is the rate of change of velocity $v$ with respect to time $t$ ($a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$).
• In Chemistry, reaction rate is the rate of change of concentration of a reactant or product with respect to time.
• In Economics, marginal cost is the rate of change of total cost with respect to the number of units produced.
• In Biology, growth rate is the rate of change of population size or biomass with respect to time.

The rate of change of $y$ with respect to $x$ at a particular value $x = x_0$ is denoted by $\left(\frac{dy}{dx}\right)_{x=x_0}$ or $f'(x_0)$. Similarly, the rate of change of $y$ with respect to $t$ at a specific time $t_0$ is denoted by $\left(\frac{dy}{dt}\right)_{t=t_0}$.

The sign of the derivative provides information about the direction of change:

• If $\frac{dy}{dx} > 0$ at a particular value of $x$, it means that $y$ is increasing as $x$ increases at that point. The function $y=f(x)$ is strictly increasing at $x$.
• If $\frac{dy}{dx} < 0$ at a particular value of $x$, it means that $y$ is decreasing as $x$ increases at that point. The function $y=f(x)$ is strictly decreasing at $x$.
• If $\frac{dy}{dx} = 0$ at a particular value of $x$, it means that $y$ is momentarily stationary (neither increasing nor decreasing) with respect to $x$ at that point. These points are often critical points where maxima or minima might occur.

The units of the rate of change $\frac{dy}{dx}$ are the units of the quantity $y$ divided by the units of the quantity $x$. For example:

• If $y$ is the area of a square in square centimetres ($\text{cm}^2$) and $x$ is the side length in centimetres ($\text{cm}$), then $\frac{dA}{dx}$ would have units of $\frac{\text{cm}^2}{\text{cm}}$, which simplifies to $\text{cm}$.
• If $y$ is the volume of water in a tank in cubic metres ($\text{m}^3$) and $t$ is time in seconds ($\text{s}$), then $\frac{dV}{dt}$ would have units of $\frac{\text{m}^3}{\text{s}}$, representing the rate of flow.
• If $y$ is the temperature in degrees Celsius ($^\circ\text{C}$) and $t$ is time in minutes, then $\frac{dT}{dt}$ would have units of $^\circ\text{C}/\text{minute}$.

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Related Rates

In numerous practical scenarios, multiple quantities are interconnected by an equation, and these quantities are all changing simultaneously with respect to time. Problems that involve finding the rate of change of one quantity based on the known rates of change of other related quantities are called related rates problems. The key to solving these problems is to recognize the relationship between the quantities and then differentiate this relationship with respect to time, typically using the Chain Rule.

Suppose quantities $u, v, w, \dots$ are related by an equation $F(u, v, w, \dots) = C$ (where $C$ is a constant) or more generally $F(u, v, w, \dots) = G(u, v, w, \dots)$. If all these quantities are changing with respect to time $t$, meaning $u=u(t), v=v(t), w=w(t), \dots$, then we can find the relationship between their rates of change $\frac{du}{dt}, \frac{dv}{dt}, \frac{dw}{dt}, \dots$ by implicitly differentiating the original equation with respect to $t$.

For instance, if $y$ is a function of $x$ ($y=f(x)$), and both $y$ and $x$ are changing with respect to time $t$ ($y=y(t), x=x(t)$), then the rates of change $\frac{dy}{dt}$ and $\frac{dx}{dt}$ are related by the Chain Rule:

This formula is fundamental to related rates problems. If the relationship involves more variables or more complex functions, the differentiation step requires careful application of differentiation rules like the product rule, quotient rule, and the Chain Rule for each term.

The general procedure for solving a related rates problem can be summarised in the following steps:

1. Understand the Problem: Read the problem carefully. Identify what is being asked for and what information is given. Determine which quantities are changing and which are constant. Drawing a diagram is often very helpful, especially for problems involving geometric shapes. Label the changing quantities with variables.

2. Assign Variables and Notation: Assign variables to all quantities that are changing with respect to time. For instance, use $r$ for radius, $A$ for area, $V$ for volume, $h$ for height, $s$ for displacement, etc. Use $t$ for time. Write down the given rates of change and the rate you need to find using derivative notation (e.g., $\frac{dr}{dt} = \dots$, find $\frac{dA}{dt}$ when $\dots$).

3. Find a Relationship: Write an equation that relates the variables you defined in the previous step. This equation must be true for all values of time $t$ within the considered interval, not just at the specific instant mentioned in the problem. This equation usually comes from geometric formulas (area, volume, Pythagorean theorem), trigonometric relationships, or other principles relevant to the problem (e.g., conservation laws).

4. Differentiate with Respect to Time: Differentiate both sides of the equation from step 3 with respect to time $t$. Remember to apply the Chain Rule whenever you differentiate a variable that is a function of $t$. For example, $\frac{d}{dt}(r^2) = 2r \frac{dr}{dt}$, $\frac{d}{dt}(V) = \frac{dV}{dt}$, $\frac{d}{dt}(x^3) = 3x^2 \frac{dx}{dt}$. Treat constants as usual (their derivative is 0).

5. Substitute and Solve: After differentiating, substitute all the known values into the resulting equation. These known values include the given rates of change and the specific values of the variables at the particular instant for which the rate is to be found. Then, solve the equation for the unknown rate of change. Make sure your final answer includes the correct units, which are determined by the units of the quantities involved (units of dependent variable per units of independent variable, which is usually time $t$).

Crucial Point: Do not substitute the specific numerical values of the variables into the equation *before* you differentiate. The relationship between the variables must be differentiated first, and only then should you plug in the instantaneous values. Substituting too early would treat the variables as constants, resulting in a derivative of zero, which is incorrect.

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Tangents and Normals

A significant geometric application of derivatives is finding the equation of the tangent line and the normal line to a curve at a given point. The derivative of a function at a specific point represents the slope of the tangent line to the graph of the function at that point.

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Slope of the Tangent

Consider a curve represented by the equation $y = f(x)$. We want to find the equation of the tangent line to this curve at a specific point $P(x_0, y_0)$ on the curve. For the point $P(x_0, y_0)$ to be on the curve, it must satisfy the equation, i.e., $y_0 = f(x_0)$.

The slope of the tangent line to the curve $y = f(x)$ at the point $P(x_0, y_0)$ is given by the value of the derivative of $f(x)$ evaluated at $x = x_0$.

Here, $\left(\frac{dy}{dx}\right)_{(x_0, y_0)}$ means that we first find the derivative $\frac{dy}{dx}$ as a function of $x$, and then substitute $x=x_0$ (and if $y_0$ appears in the derivative expression, substitute $y=y_0$) to get the numerical value of the slope at that specific point.

Equation of the Tangent

Once we have the slope of the tangent line $m_T$ at the point $(x_0, y_0)$, we can find the equation of the tangent line using the point-slope form of the equation of a line, which is $(y - y_1) = m(x - x_1)$.

Using the point $P(x_0, y_0)$ and the slope $m_T = \left(\frac{dy}{dx}\right)_{(x_0, y_0)}$, the equation of the tangent line is:

Substituting the expression for $m_T$:

This is the equation of the tangent to the curve $y = f(x)$ at the point $(x_0, y_0)$.

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Slope of the Normal

The normal to a curve at a given point is defined as the line that is perpendicular to the tangent line at that same point.

We know from coordinate geometry that if two non-vertical lines are perpendicular, the product of their slopes is -1. Let $m_T$ be the slope of the tangent and $m_N$ be the slope of the normal at the point $(x_0, y_0)$.

If $m_T$ is finite and non-zero, then:

Therefore, the slope of the normal line $m_N$ is given by:

Equation of the Normal

Similar to the tangent, we can find the equation of the normal line using the point-slope form, with the point $P(x_0, y_0)$ and the slope of the normal $m_N = -\frac{1}{\left(\frac{dy}{dx}\right)_{(x_0, y_0)}}$ (provided $\left(\frac{dy}{dx}\right)_{(x_0, y_0)} \neq 0$).

The equation of the normal line is:

Substituting the expression for $m_N$:

This is the equation of the normal to the curve $y = f(x)$ at the point $(x_0, y_0)$.

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Special Cases

We need to consider cases where the slope of the tangent is zero or undefined.

1.

Horizontal Tangent:

If the tangent line at $(x_0, y_0)$ is horizontal, its slope is 0. This occurs when $\left(\frac{dy}{dx}\right)_{(x_0, y_0)} = 0$. The equation of a horizontal line passing through $(x_0, y_0)$ is $y = y_0$.

In this case, the normal line, being perpendicular to a horizontal line, must be a vertical line. A vertical line passing through $(x_0, y_0)$ has the equation $x = x_0$. Note that the slope of a vertical line is undefined, which is consistent with $m_N = -1/m_T$ when $m_T = 0$.

2.

Vertical Tangent:

If the tangent line at $(x_0, y_0)$ is vertical, its slope is undefined. This typically occurs when $\left(\frac{dy}{dx}\right)_{(x_0, y_0)}$ approaches infinity. In many cases, this happens when $\left(\frac{dx}{dy}\right)_{(x_0, y_0)} = 0$ (if we consider $x$ as a function of $y$, $x=g(y)$). The equation of a vertical line passing through $(x_0, y_0)$ is $x = x_0$.

In this case, the normal line, being perpendicular to a vertical line, must be a horizontal line. A horizontal line passing through $(x_0, y_0)$ has the equation $y = y_0$. Note that the slope of a horizontal line is 0.

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Approximations, Errors and Differentials

Derivatives provide a powerful tool for approximating the change in the value of a function when there is a small change in its independent variable. This concept is closely tied to the idea of differentials and their use in estimating errors in measurements and calculations.

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Differentials

Let $y = f(x)$ be a differentiable function of $x$. We introduce the concept of differentials $dx$ and $dy$.

The differential of $x$, denoted by $dx$, is defined as an independent variable. It is usually taken to be equal to a small change in $x$, which is denoted by $\Delta x$.

The differential of $y$, denoted by $dy$, is defined in terms of $dx$ and the derivative of the function $f'(x) = \frac{dy}{dx}$.

From the definition of the derivative, we know that $\frac{dy}{dx} = \lim\limits_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$. For small values of $\Delta x$, the ratio $\frac{\Delta y}{\Delta x}$ is approximately equal to the derivative $f'(x)$.

Multiplying both sides by $\Delta x$:

Comparing this with the definition of $dy$ from equation (2) and using $dx = \Delta x$ from equation (1), we see that for small changes in $x$:

Here, $\Delta y = f(x + \Delta x) - f(x)$ represents the actual change in the value of the function $y$ when $x$ changes by $\Delta x$. The differential $dy$ represents the approximate change in the value of the function, calculated using the tangent line at $x$. Geometrically, $dy$ is the change in the y-coordinate along the tangent line when $x$ changes by $dx$, while $\Delta y$ is the change in the y-coordinate along the curve itself. For small $\Delta x$, the tangent line is a good approximation to the curve, hence $dy \approx \Delta y$.

From the approximation $\Delta y \approx dy$, we have:

Rearranging this equation gives the formula for linear approximation:

This formula is the basis for using differentials to approximate function values and estimate errors.

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Approximation and Errors

The formula $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$ is particularly useful for approximating values of functions at points slightly away from a point where the function value and its derivative are easily known. For example, it can be used to approximate values like $\sqrt{26}$, $(8.01)^{1/3}$, $\sin(31^\circ)$, etc.

In the context of error estimation, if $\Delta x$ (or $dx$) represents a small error made in the measurement of the independent variable $x$, then the corresponding approximate error in the calculated value of the dependent variable $y = f(x)$ is given by $dy$.

Let $\delta x$ be the small error in the measurement of $x$. This error $\delta x$ is equivalent to $\Delta x$ or $dx$. The approximate error in the value of $y$ is $\delta y$, which is approximated by $dy$.

Different types of errors are defined as follows:

• Absolute Error: The magnitude of the error.
  • Absolute error in $x$: $|\Delta x|$ or $|dx|$
  • Approximate absolute error in $y$: $|\Delta y| \approx |dy| = |f'(x) dx|$
• Relative Error: The ratio of the absolute error to the actual value of the quantity. For small errors, we use the approximate value.
  • Relative error in $x$: $\frac{\Delta x}{x}$ or $\frac{dx}{x}$
  • Approximate relative error in $y$: $\frac{\Delta y}{y} \approx \frac{dy}{y} = \frac{f'(x) dx}{f(x)}$
• Percentage Error: The relative error multiplied by 100%.
  • Percentage error in $x$: $\left(\frac{\Delta x}{x}\right) \times 100\%$ or $\left(\frac{dx}{x}\right) \times 100\%$
  • Approximate percentage error in $y$: $\left(\frac{dy}{y}\right) \times 100\% = \left(\frac{f'(x) dx}{f(x)}\right) \times 100\%$

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Increasing and Decreasing Functions

The derivative of a function is a powerful tool that allows us to analyse the behaviour of the function, specifically whether it is increasing, decreasing, or constant over a particular interval. This property is known as the monotonicity of the function. Understanding monotonicity is essential for sketching the graph of a function and determining its local and global extreme values.

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Definitions

Let $f$ be a real-valued function defined on an interval $I$. The interval $I$ can be any type of interval (open, closed, semi-open, finite, or infinite).

1. $f$ is said to be increasing on $I$ if for any two points $x_1, x_2$ in $I$, whenever $x_1 < x_2$, we have $f(x_1) \leq f(x_2)$. This means the function value does not decrease as $x$ increases.

2. $f$ is said to be strictly increasing on $I$ if for any two points $x_1, x_2$ in $I$, whenever $x_1 < x_2$, we have $f(x_1) < f(x_2)$. This means the function value always increases as $x$ increases.

3. $f$ is said to be decreasing on $I$ if for any two points $x_1, x_2$ in $I$, whenever $x_1 < x_2$, we have $f(x_1) \geq f(x_2)$. This means the function value does not increase as $x$ increases.

4. $f$ is said to be strictly decreasing on $I$ if for any two points $x_1, x_2$ in $I$, whenever $x_1 < x_2$, we have $f(x_1) > f(x_2)$. This means the function value always decreases as $x$ increases.

5. $f$ is said to be constant on $I$ if for all $x_1, x_2$ in $I$, $f(x_1) = f(x_2)$.

A function is said to be monotonic on an interval if it is either increasing or decreasing on that interval. It is strictly monotonic if it is either strictly increasing or strictly decreasing.

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Criterion for Increasing and Decreasing Functions (First Derivative Test for Monotonicity)

The sign of the first derivative of a function provides a simple criterion to determine its monotonicity on an interval.

Let $f$ be a function that is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$.

1. If $f'(x) > 0$ for all $x \in (a, b)$, then $f$ is strictly increasing on $[a, b]$.

2. If $f'(x) < 0$ for all $x \in (a, b)$, then $f$ is strictly decreasing on $[a, b]$.

3. If $f'(x) = 0$ for all $x \in (a, b)$, then $f$ is constant on $[a, b]$.

Note on Non-Strict Monotonicity:

• If $f'(x) \geq 0$ for all $x \in (a, b)$, then $f$ is increasing on $[a, b]$.
• If $f'(x) \leq 0$ for all $x \in (a, b)$, then $f$ is decreasing on $[a, b]$.

In these cases, $f'(x)$ can be zero at some points in the interval, but if $f'(x)=0$ only at isolated points (not over an entire sub-interval), the function is still strictly monotonic on the interval where $f'(x) > 0$ or $f'(x) < 0$. However, if $f'(x) = 0$ over an entire sub-interval, the function is constant over that sub-interval. The criteria $f'(x) > 0$ and $f'(x) < 0$ guarantee strict monotonicity.

Proof (using Mean Value Theorem)

Let $x_1, x_2$ be any two points in the interval $I \subseteq [a, b]$ such that $x_1 < x_2$. Since $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, it satisfies the conditions of the Mean Value Theorem (MVT) on the sub-interval $[x_1, x_2]$.

According to the Mean Value Theorem, there exists a point $c \in (x_1, x_2)$ such that:

Since $x_1 < x_2$, we have $x_2 - x_1 > 0$. The sign of $f(x_2) - f(x_1)$ (which determines whether $f(x_2) > f(x_1)$, $f(x_2) < f(x_1)$, or $f(x_2) = f(x_1)$) depends directly on the sign of $f'(c)$.

Case 1: If $f'(x) > 0$ for all $x \in (a, b)$, then in particular, $f'(c) > 0$. From the MVT equation:

Since $x_2 - x_1 > 0$, this inequality implies $f(x_2) - f(x_1) > 0$, which means $f(x_2) > f(x_1)$. Since this holds for any $x_1 < x_2$ in $I$, $f$ is strictly increasing on $I$. Since $I$ was any sub-interval of $[a, b]$, $f$ is strictly increasing on $[a, b]$.

Case 2: If $f'(x) < 0$ for all $x \in (a, b)$, then $f'(c) < 0$. From the MVT equation:

Since $x_2 - x_1 > 0$, this inequality implies $f(x_2) - f(x_1) < 0$, which means $f(x_2) < f(x_1)$. Since this holds for any $x_1 < x_2$ in $I$, $f$ is strictly decreasing on $I$, and hence on $[a, b]$.

Case 3: If $f'(x) = 0$ for all $x \in (a, b)$, then $f'(c) = 0$. From the MVT equation:

Since $x_2 - x_1 \neq 0$, this implies $f(x_2) - f(x_1) = 0$, which means $f(x_2) = f(x_1)$. Since this holds for any $x_1, x_2$ in $I$, $f$ is constant on $I$, and hence on $[a, b]$.

The proofs for $f'(x) \geq 0 \implies f$ is increasing and $f'(x) \leq 0 \implies f$ is decreasing follow similarly, considering the possibility of $f'(c)=0$.

Procedure to Find Intervals of Monotonicity

To determine the intervals where a function $f(x)$ is strictly increasing or strictly decreasing:

1. Find the derivative of the function, $f'(x)$.

2. Find the critical points of $f(x)$. Critical points are the values of $x$ in the domain of $f$ where $f'(x) = 0$ or $f'(x)$ is undefined. These points are potential turning points where the function's monotonicity might change.

3. Use the critical points to divide the domain of the function into disjoint intervals. These intervals are the candidates for intervals of strict monotonicity.

4. Choose a test value (any point) $c$ within each interval. Evaluate the sign of $f'(c)$ at this test value.

• If $f'(c) > 0$, then $f'(x) > 0$ for all $x$ in that interval (assuming $f'$ is continuous, which is usually the case for functions studied at this level, except possibly at the critical points themselves). Therefore, the function $f$ is strictly increasing on that interval.
• If $f'(c) < 0$, then $f'(x) < 0$ for all $x$ in that interval. Therefore, the function $f$ is strictly decreasing on that interval.

5. State the intervals where $f$ is strictly increasing and strictly decreasing. If the function is continuous at the endpoints of the intervals derived from the critical points, these endpoints are usually included in the intervals of (non-strict) increasing or decreasing, but for strict monotonicity, open intervals are typically used unless specified otherwise or based on convention. In board exams, intervals are usually written including endpoints if the function is continuous there.

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Maxima and Minima

Finding the extreme values of a function, i.e., its highest and lowest points, is a fundamental problem in calculus and has numerous applications in optimisation. Derivatives play a crucial role in locating these values. The extreme values can be classified as absolute (global) or local (relative).

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Absolute Maximum and Minimum Values (Global Extrema)

The absolute maximum or minimum value of a function over a given interval is the largest or smallest value that the function attains anywhere in that entire interval.

Let $f$ be a function defined on an interval $I$.

• A function $f$ is said to have an absolute maximum value on $I$ at a point $c \in I$ if $f(c) \geq f(x)$ for all $x$ in the interval $I$. The value $f(c)$ is the absolute maximum value of $f$ on $I$, and $c$ is the point of absolute maximum.
• A function $f$ is said to have an absolute minimum value on $I$ at a point $c \in I$ if $f(c) \leq f(x)$ for all $x$ in the interval $I$. The value $f(c)$ is the absolute minimum value of $f$ on $I$, and $c$ is the point of absolute minimum.

The absolute maximum and minimum values are collectively called the **global extrema** or **absolute extrema** of the function on the interval $I$. A function may or may not have absolute extrema on an open interval or the entire real line. However, a key theorem guarantees their existence under certain conditions:

Extreme Value Theorem:

If a function $f$ is continuous on a closed interval $[a, b]$, then $f$ is guaranteed to attain both an absolute maximum value and an absolute minimum value at some points within that interval $[a, b]$.

For a continuous function on a closed interval, these absolute extreme values must occur either at the endpoints of the interval ($x=a$ or $x=b$) or at the critical points of the function within the open interval $(a, b)$.

Critical Point

A critical point (or critical number) of a function $f$ is a point $c$ in the domain of $f$ where either:

• $f'(c) = 0$ (the tangent line is horizontal), OR
• $f'(c)$ is undefined.

Critical points are candidates for local (and sometimes global) maxima or minima because the function's behaviour (increasing/decreasing) can change at these points.

Procedure for Finding Absolute Maximum and Minimum Values on a Closed Interval $[a, b]$

If a function $f$ is continuous on a closed interval $[a, b]$, the following systematic approach can be used to find its absolute maximum and minimum values:

1. Find all critical points of the function $f(x)$ that lie within the open interval $(a, b)$. To do this, compute the derivative $f'(x)$, set $f'(x) = 0$ and solve for $x$, and also identify any points in $(a, b)$ where $f'(x)$ is undefined but $f(x)$ is defined.

2. Evaluate the function $f(x)$ at all the critical points found in step 1.

3. Evaluate the function $f(x)$ at the endpoints of the interval, i.e., find $f(a)$ and $f(b)$.

4. Compare all the function values calculated in steps 2 and 3.

• The largest among these values is the absolute maximum value of $f$ on $[a, b]$.
• The smallest among these values is the absolute minimum value of $f$ on $[a, b]$.

The point(s) where the absolute maximum value occurs is/are the point(s) of absolute maximum, and similarly for the absolute minimum.

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Local Maximum and Local Minimum Values

Beyond finding the absolute highest and lowest points on an interval, derivatives also help us identify local (or relative) extrema. These are points where the function reaches a peak or a valley in its immediate neighbourhood, even if there are higher or lower points elsewhere in the domain.

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Definitions

Let $f$ be a function defined on an interval $I$, and let $c$ be an interior point of $I$ (a point that is not an endpoint).

• A function $f$ is said to have a local maximum value at $c$ if there exists at least one open interval $(c - \delta, c + \delta)$, where $\delta > 0$ is a small positive number, such that $(c - \delta, c + \delta) \subset I$ and $f(c) \geq f(x)$ for all $x$ in $(c - \delta, c + \delta)$. The value $f(c)$ is called a local maximum value of $f$. This means $f(c)$ is the highest function value in some neighbourhood around $c$.
• A function $f$ is said to have a local minimum value at $c$ if there exists at least one open interval $(c - \delta, c + \delta)$, where $\delta > 0$, such that $(c - \delta, c + \delta) \subset I$ and $f(c) \leq f(x)$ for all $x$ in $(c - \delta, c + \delta)$. The value $f(c)$ is called a local minimum value of $f$. This means $f(c)$ is the lowest function value in some neighbourhood around $c$.

Points where a function has a local maximum or a local minimum are collectively called local extrema or relative extrema. A point $c$ where a local extremum occurs must be a critical point of the function (where $f'(c) = 0$ or $f'(c)$ is undefined, provided $c$ is in the domain). This is a necessary condition, but not sufficient (i.e., not all critical points are local extrema).

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Tests for Local Extrema

To determine whether a critical point corresponds to a local maximum, a local minimum, or neither, we use tests based on the derivative(s) of the function.

First Derivative Test

The First Derivative Test relies on examining the sign of the first derivative, $f'(x)$, as $x$ moves across a critical point. The sign of the derivative tells us whether the function is increasing or decreasing in the vicinity of the critical point.

Let $c$ be a critical point of a continuous function $f$ defined on an interval $I$. Suppose $f$ is differentiable in $(c-\delta, c)$ and $(c, c+\delta)$ for some $\delta > 0$.

• Condition for Local Maximum: If $f'(x)$ changes sign from positive to negative as $x$ increases through $c$, then $f$ has a local maximum at $c$. Graphically, the function increases up to $c$ and then decreases after $c$, forming a peak.

Sign of $f'(x)$ for $x < c$: Positive ($f$ increasing)

Sign of $f'(x)$ for $x > c$: Negative ($f$ decreasing)

$\implies$ Local Maximum at $c$.

• Condition for Local Minimum: If $f'(x)$ changes sign from negative to positive as $x$ increases through $c$, then $f$ has a local minimum at $c$. Graphically, the function decreases up to $c$ and then increases after $c$, forming a valley.

Sign of $f'(x)$ for $x < c$: Negative ($f$ decreasing)

Sign of $f'(x)$ for $x > c$: Positive ($f$ increasing)

$\implies$ Local Minimum at $c$.

• Condition for Neither Maximum nor Minimum: If $f'(x)$ does not change sign as $x$ increases through $c$ (i.e., $f'(x)$ is positive on both sides of $c$ or negative on both sides), then $c$ is neither a local maximum nor a local minimum. In such cases, the function continues to increase or decrease through the critical point. (Note: While $f'(c)=0$ or is undefined, $c$ might be a point of inflection, related to concavity change, but the First Derivative Test only tells us about monotonicity change).

Sign of $f'(x)$ for $x < c$: Positive

Sign of $f'(x)$ for $x > c$: Positive

$\implies$ Neither Local Max nor Min at $c$ (possibly point of inflection).

Sign of $f'(x)$ for $x < c$: Negative

Sign of $f'(x)$ for $x > c$: Negative

$\implies$ Neither Local Max nor Min at $c$ (possibly point of inflection).

Procedure using First Derivative Test:

1. Find the derivative $f'(x)$.

2. Find the critical points of $f(x)$ (where $f'(x)=0$ or $f'(x)$ is undefined) that are in the domain of $f$.

3. Arrange the critical points in increasing order on a number line. These points divide the domain into disjoint intervals.

4. For each critical point $c$, examine the sign of $f'(x)$ in an open interval immediately to the left of $c$ and in an open interval immediately to the right of $c$. This can be done by picking a test point in each interval and evaluating $f'$ at that point.

5. Apply the sign change conditions of the First Derivative Test to classify each critical point as a local maximum, local minimum, or neither.

6. If a local extremum exists at $c$, calculate the local extreme value by finding $f(c)$.

Second Derivative Test

The Second Derivative Test uses the value of the second derivative at a critical point where the first derivative is zero ($f'(c)=0$). It is often quicker when applicable, but it requires the existence of the second derivative.

Let $f$ be a function that is differentiable on an open interval $I$, and let $c$ be an interior point in $I$ such that $f'(c) = 0$. Assume the second derivative $f''(c)$ exists at $c$.

• Condition for Local Maximum: If $f''(c) < 0$, then $f$ has a local maximum at $c$. (The function's graph is concave down at $c$).
• Condition for Local Minimum: If $f''(c) > 0$, then $f$ has a local minimum at $c$. (The function's graph is concave up at $c$).
• When the Test Fails: If $f''(c) = 0$ or $f''(c)$ is undefined, the Second Derivative Test gives no conclusion about local extrema at $c$. In such cases, you must use the First Derivative Test to determine the nature of the critical point.

Procedure using Second Derivative Test:

1. Find the first derivative $f'(x)$ and the second derivative $f''(x)$.

2. Find the critical points $c$ by setting $f'(x) = 0$ and solving for $x$. This test *only* applies to critical points where $f'(c)=0$.

3. For each critical point $c$ found in step 2, evaluate the second derivative $f''(c)$.

4. Apply the conditions based on the sign of $f''(c)$:

• If $f''(c) < 0$, $f$ has a local maximum at $x=c$. The local maximum value is $f(c)$.
• If $f''(c) > 0$, $f$ has a local minimum at $x=c$. The local minimum value is $f(c)$.
• If $f''(c) = 0$ or $f''(c)$ is undefined, the test fails. Proceed to use the First Derivative Test for this specific critical point.

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Practical Problems on Maxima and Minima

One of the most significant and practical applications of differential calculus is in solving optimization problems. These problems involve finding the maximum or minimum value of a certain quantity (like area, volume, profit, cost, time, distance, etc.) under given constraints. The ability to model such situations mathematically and then use derivatives to find the extreme values makes calculus an indispensable tool in various fields like engineering, economics, physics, and business. Solving these problems typically involves identifying the quantity to be optimized, expressing it as a function, and then finding the extrema of that function using the techniques discussed earlier (finding critical points, and applying the First or Second Derivative Test).

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Steps to Solve Optimization Problems

Solving optimization problems using calculus generally follows a systematic approach:

1. Understand the Problem: Read the problem statement carefully to fully grasp what is being asked. Identify the quantity that needs to be maximised or minimised. Assign a symbol (variable) to this quantity, say $Q$. Identify all other relevant quantities and assign variables to them as well.

2. Draw a Diagram (if applicable): For geometric problems, drawing a clear diagram and labelling the relevant variables can greatly help in visualising the relationships between the quantities.

3. Find Relationships and Formulate the Function: Write down any equations that relate the variables identified in the problem. These usually come from the problem description (e.g., fixed sum, fixed perimeter, formulas for area/volume, etc.). Use these relationships to express the quantity $Q$ (which you want to optimize) as a function of a single independent variable, say $x$. For example, if $Q$ depends on $x$ and $y$, use a relation between $x$ and $y$ to express $y$ in terms of $x$ and substitute it into the expression for $Q$, resulting in $Q(x)$.

4. Determine the Domain: Identify the feasible range of the independent variable $x$ based on the constraints of the problem. For instance, lengths, widths, radii, etc., must typically be non-negative. This will define the domain (an interval) for the function $Q(x)$. This domain might be open or closed depending on the problem.

5. Find Critical Points: Calculate the first derivative of the function $Q(x)$, i.e., $Q'(x)$. Find the critical points within the determined domain by setting $Q'(x) = 0$ or by finding points in the domain where $Q'(x)$ is undefined.

6. Test Critical Points: Use either the First Derivative Test or the Second Derivative Test to classify each critical point found in step 5 as a local maximum, a local minimum, or neither. The First Derivative Test is generally more robust as it works even when the second derivative is zero or undefined. The Second Derivative Test is often quicker if the second derivative is easy to compute and is non-zero at the critical points.

7. Evaluate at Endpoints (if applicable): If the domain determined in step 4 is a closed interval $[a, b]$ and you are looking for absolute extrema, you must also evaluate the function $Q(x)$ at the endpoints $a$ and $b$, in addition to evaluating it at the critical points within $(a, b)$.

8. Identify the Optimal Value and Answer the Question: Based on the results of the tests in step 6 or the comparison of values in step 7 (for closed intervals), identify the value(s) of $x$ that yield the required maximum or minimum value of $Q$. State the answer clearly, providing both the value(s) of the independent variable(s) and the resulting maximum or minimum value of the quantity $Q$, ensuring units are included if applicable.

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