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Rolle’s Theorem: Statement and Conditions Geometric Interpretation of Rolle's Theorem Lagranges’s Mean Value Theorem: Statement and Conditions
Geometric Interpretation of Lagrange's Mean Value Theorem Verification and Applications of Mean Value Theorems


Mean Value Theorems



Rolle’s Theorem: Statement and Conditions

Rolle's Theorem is a fundamental result in differential calculus. It is a special case of the Mean Value Theorem and provides a condition under which a function's derivative must be zero at some point within an interval. This point corresponds to a horizontal tangent on the graph of the function.

Statement of Rolle's Theorem

Let $f$ be a real-valued function defined on the closed interval $[a, b]$. Rolle's Theorem states that if $f$ satisfies the following three conditions:

  1. Continuity on the closed interval $[a, b]$: The function $f$ must be continuous at every point in the interval $[a, b]$, including the endpoints $a$ and $b$. This means the graph of the function must be an unbroken curve from the point $(a, f(a))$ to $(b, f(b))$.

  2. Differentiability on the open interval $(a, b)$: The function $f$ must be differentiable at every point in the open interval $(a, b)$, i.e., for all $x$ such that $a < x < b$. This means the graph of the function must be smooth within the interval, without any sharp corners, cusps, or vertical tangents.

  3. Equal function values at the endpoints: The value of the function at the left endpoint $a$ must be equal to the value of the function at the right endpoint $b$.

    "$f(a) = f(b)$"

    ... (i)

Conclusion: If all three of these conditions are met, then there exists at least one number $c$ in the open interval $(a, b)$ (meaning $a < c < b$) such that the derivative of the function at $c$ is zero:

$f'(c) = 0$

Rolle's theorem guarantees the *existence* of such a point $c$, but it does not provide a method for finding the value of $c$. It also states that there is *at least one* such point, meaning there could be more than one point in the interval where the derivative is zero.


Explanation of Conditions and Importance

Each of the three conditions in Rolle's Theorem is essential. If any one of them is not satisfied, the conclusion ($f'(c) = 0$ for some $c \in (a, b)$) may not hold.

Rolle's Theorem is important theoretically as it is used to prove the Mean Value Theorem, which is one of the most significant theorems in calculus. It also has applications in showing the existence of roots of derivatives and separating roots of functions.



Geometric Interpretation of Rolle's Theorem

The geometric interpretation of Rolle's Theorem provides a visual understanding of its conclusion.

Consider the graph of a function $y = f(x)$ that satisfies the conditions of Rolle's Theorem on the interval $[a, b]$.

The condition $f(a) = f(b)$ means that the starting point $(a, f(a))$ and the ending point $(b, f(b))$ of the curve segment have the same y-coordinate (are at the same height).

The conditions of continuity on $[a, b]$ and differentiability on $(a, b)$ mean that the curve between these two points is unbroken and smooth (no sharp corners or cusps).

If the function starts and ends at the same height, and the curve is smooth and continuous in between, then the function must change direction at least once to get from $(a, f(a))$ to $(b, f(b))$. If the function increases from $(a, f(a))$, it must eventually decrease to get back to the height $f(b)$. If it decreases, it must eventually increase. These changes in direction typically occur at local maximum or local minimum points within the interval $(a, b)$.

At such a local maximum or minimum point $(c, f(c))$ in the open interval $(a, b)$, where the function is differentiable, the tangent line to the curve must be horizontal. The slope of a horizontal line is zero.

The slope of the tangent line at any point $x=c$ is given by the derivative $f'(c)$.

Therefore, the conclusion of Rolle's Theorem, $f'(c) = 0$ for some $c \in (a, b)$, means that there exists at least one point $c$ strictly between $a$ and $b$ where the tangent line to the graph of $f(x)$ is horizontal.

Geometric Interpretation of Rolle's Theorem

The diagram shows examples where there is one such point (at a local max or min) or multiple such points (alternating local max and min). Rolle's Theorem guarantees that there will be at least one horizontal tangent between $a$ and $b$ if the conditions are met.



Lagranges’s Mean Value Theorem: Statement and Conditions

Lagrange's Mean Value Theorem (MVT) is a pivotal theorem in calculus that extends Rolle's Theorem. It relates the average rate of change of a function over an interval to its instantaneous rate of change at some point within that interval.

Statement of Lagrange's Mean Value Theorem

Let $f$ be a real-valued function defined on the closed interval $[a, b]$. The Mean Value Theorem states that if $f$ satisfies the following two conditions:

  1. Continuity on the closed interval $[a, b]$: The function $f$ must be continuous at every point in $[a, b]$, including $a$ and $b$. The graph is an unbroken curve from $(a, f(a))$ to $(b, f(b))$.

  2. Differentiability on the open interval $(a, b)$: The function $f$ must be differentiable at every point in the open interval $(a, b)$, i.e., for all $x$ such that $a < x < b$. The graph is smooth without sharp corners, cusps, or vertical tangents strictly between $a$ and $b$.

Conclusion: If these two conditions are met, then there exists at least one number $c$ in the open interval $(a, b)$ (meaning $a < c < b$) such that the derivative of the function at $c$ is equal to the average rate of change of the function over the entire interval $[a, b]$. The average rate of change is given by the slope of the secant line connecting the endpoints $(a, f(a))$ and $(b, f(b))$.

$f'(c) = \frac{f(b) - f(a)}{b - a}$

The MVT guarantees the existence of such a point $c$ in $(a, b)$, but it does not provide a method to find it. The value $c$ is often referred to as a "mean value" because $f'(c)$ equals the average rate of change.


Relation of MVT to Rolle's Theorem

Rolle's Theorem can be seen as a direct consequence or a special case of Lagrange's Mean Value Theorem. The conditions for Rolle's Theorem are the two conditions for the MVT plus one additional condition: $f(a) = f(b)$.

If the third condition of Rolle's Theorem, $f(a) = f(b)$, is added to the conditions of the Mean Value Theorem, the formula for the average rate of change on the right side of the MVT conclusion simplifies:

Average Rate of Change $= \frac{f(b) - f(a)}{b - a}$

If $f(a) = f(b)$, then $f(b) - f(a) = 0$. Assuming $a \neq b$, the denominator $b-a \neq 0$.

Average Rate of Change $= \frac{0}{b - a} = 0$

[If $f(a) = f(b)$ and $a \neq b$]

Substituting this into the conclusion of the MVT, $f'(c) = \frac{f(b) - f(a)}{b - a}$, we get:

"$f'(c) = 0$"

This is exactly the conclusion of Rolle's Theorem. Thus, if the conditions of Rolle's Theorem are satisfied, the Mean Value Theorem guarantees the existence of a point $c$ where the derivative is equal to the average rate of change (which is 0 because the endpoint values are equal).

Rolle's theorem pinpoints the existence of horizontal tangents when $f(a)=f(b)$, while the MVT generalizes this to say that the slope of the tangent at some point equals the slope of the secant connecting the endpoints, regardless of whether $f(a)$ and $f(b)$ are equal.


Example 1. Verify that the function $f(x) = x^2$ satisfies the conditions of the Mean Value Theorem on the interval $[0, 2]$. Find all values of $c$ in the open interval $(0, 2)$ that satisfy the conclusion of the theorem.

Answer:

We need to verify the conditions of the Mean Value Theorem for $f(x) = x^2$ on the closed interval $[a, b] = [0, 2]$, and then find the value(s) of $c$ in the open interval $(0, 2)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.

Verification of Conditions:

The conditions for the Mean Value Theorem are Continuity on $[a, b]$ and Differentiability on $(a, b)$.

1. Continuity on $[0, 2]$: The function $f(x) = x^2$ is a polynomial function. Polynomial functions are continuous for all real numbers. Therefore, $f(x) = x^2$ is continuous on the closed interval $[0, 2]$. Condition 1 is satisfied.

2. Differentiability on $(0, 2)$: The derivative of $f(x) = x^2$ is $f'(x) = 2x$ (using the Power Rule). This derivative exists for all real numbers. Therefore, $f(x) = x^2$ is differentiable on the open interval $(0, 2)$. Condition 2 is satisfied.

Since both conditions are satisfied, the Mean Value Theorem applies to $f(x) = x^2$ on the interval $[0, 2]$.

Finding the value(s) of $c$:

The conclusion of the Mean Value Theorem states that there exists at least one $c \in (0, 2)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.

First, calculate the average rate of change (the right side of the equation).

Here $a=0$ and $b=2$.

$f(a) = f(0) = 0^2 = 0$.

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

Average Rate of Change $= \frac{f(2) - f(0)}{2 - 0} = \frac{4 - 0}{2 - 0} = \frac{4}{2} = 2$

Next, find the derivative of the function, which is $f'(x) = 2x$.

Now, set $f'(c)$ equal to the average rate of change and solve for $c$:

"$f'(c) = 2$"

"$2c = 2$"

[Substitute $f'(c)=2c$]

Solve for $c$:

"$c = 1$"

We must check if this value of $c$ is in the open interval $(0, 2)$. The value $c=1$ is indeed in the interval $(0, 2)$.

The Mean Value Theorem guarantees at least one such value, and in this case, we found exactly one value $c=1$.

The value of $c$ that satisfies the conclusion of the theorem is $c = 1$.



Geometric Interpretation of Lagrange's Mean Value Theorem

The geometric interpretation of the Mean Value Theorem provides a visual analogy that clarifies its meaning and relationship between the average rate of change and the instantaneous rate of change.

Consider the graph of a function $y = f(x)$ that is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$. Let $A$ be the point $(a, f(a))$ and $B$ be the point $(b, f(b))$ on the graph.

The conclusion of the Mean Value Theorem, $f'(c) = \frac{f(b) - f(a)}{b - a}$ for some $c \in (a, b)$, means that there exists at least one point $c$ strictly between $a$ and $b$ such that the slope of the tangent line to the curve at $(c, f(c))$ is exactly equal to the slope of the secant line connecting the endpoints $(a, f(a))$ and $(b, f(b))$.

Geometrically, this means that the tangent line to the curve at $x=c$ is parallel to the secant line connecting the endpoints of the interval.

Geometric Interpretation of Lagrange's Mean Value Theorem

The diagram illustrates that there can be one or more such points $c$ in the open interval $(a, b)$. The MVT guarantees the existence of at least one such point, but not necessarily its uniqueness.

The geometric interpretation makes it clear how the MVT generalizes Rolle's Theorem. If $f(a) = f(b)$, the secant line connecting $(a, f(a))$ and $(b, f(b))$ is horizontal (slope is 0). In this special case, the MVT guarantees a point $c$ where the tangent line is parallel to the horizontal secant, meaning the tangent line itself is horizontal ($f'(c) = 0$), which is the conclusion of Rolle's Theorem.


Example 1. Verify the Mean Value Theorem for the function $f(x) = x^2$ on the interval $[0, 2]$ and find a value of $c$ that satisfies the conclusion of the theorem.

Answer:

We need to verify the conditions of the Mean Value Theorem for $f(x) = x^2$ on the closed interval $[a, b] = [0, 2]$, and then find the value(s) of $c$ in the open interval $(0, 2)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.

Verification of Conditions:

The conditions for the Mean Value Theorem are Continuity on $[a, b]$ and Differentiability on $(a, b)$.

1. Continuity on $[0, 2]$: The function $f(x) = x^2$ is a polynomial function. Polynomial functions are continuous for all real numbers. Therefore, $f(x) = x^2$ is continuous on the closed interval $[0, 2]$. Condition 1 is satisfied.

2. Differentiability on $(0, 2)$: The derivative of $f(x) = x^2$ is $f'(x) = 2x$ (using the Power Rule). This derivative exists for all real numbers. Therefore, $f(x) = x^2$ is differentiable on the open interval $(0, 2)$. Condition 2 is satisfied.

Since both conditions are satisfied, the Mean Value Theorem applies to $f(x) = x^2$ on the interval $[0, 2]$.

Finding the value(s) of $c$:

The conclusion of the Mean Value Theorem states that there exists at least one $c \in (0, 2)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.

First, calculate the average rate of change (the right side of the equation), which is the slope of the secant line connecting $(0, f(0))$ and $(2, f(2))$.

Here $a=0$ and $b=2$.

$f(a) = f(0) = 0^2 = 0$.

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

Average Rate of Change $= \frac{f(2) - f(0)}{2 - 0} = \frac{4 - 0}{2 - 0} = \frac{4}{2} = 2$

Next, find the derivative of the function, which represents the instantaneous rate of change at any point $x$.

$f'(x) = \frac{d}{dx}(x^2) = 2x$.

Now, set the instantaneous rate of change at $c$, $f'(c) = 2c$, equal to the average rate of change (which is 2) and solve for $c$:

"$f'(c) = 2$"

[Equating instantaneous and average rates]

"$2c = 2$"

[Substitute $f'(c)=2c$]

Solve for $c$:

"$c = \frac{2}{2} = 1$"

We must check if this value of $c$ lies in the open interval $(0, 2)$. The value $c=1$ is indeed strictly between 0 and 2.

The Mean Value Theorem guarantees at least one such value exists, and we found exactly one value $c=1$. At $x=1$, the tangent line to $y=x^2$ has a slope of $f'(1)=2$, which is the same as the slope of the secant line connecting $(0,0)$ and $(2,4)$.

The value of $c$ that satisfies the conclusion of the theorem is $c = 1$.



Verification and Applications of Mean Value Theorems

The Mean Value Theorems (Rolle's Theorem and Lagrange's Mean Value Theorem) are powerful existence theorems. When applying them to a specific function on a specific interval, the process involves verifying that the function and interval satisfy the conditions of the theorem and, if required, finding the value(s) of $c$ whose existence is guaranteed by the conclusion.

Verification of Mean Value Theorems

To verify whether a given function $f(x)$ satisfies the conditions of either Rolle's Theorem or the Mean Value Theorem on a closed interval $[a, b]$, and to find the value(s) of $c$ guaranteed by the conclusion, follow these steps:

  1. Check the Conditions of the Theorem:
    • Continuity: Determine if the function $f(x)$ is continuous on the closed interval $[a, b]$. Recall that polynomial functions, exponential functions ($a^x, e^x$), sine ($\sin x$), and cosine ($\cos x$) are continuous everywhere. Rational functions are continuous on their domain (everywhere except where the denominator is zero). Logarithmic functions ($\log_a x, \ln x$) are continuous on their domain $(0, \infty)$. Pay attention to potential discontinuities within or at the endpoints of the interval, such as points where the function is undefined or piecewise function definitions change.
    • Differentiability: Determine if the function $f(x)$ is differentiable on the open interval $(a, b)$. This requires finding the derivative $f'(x)$ and checking if it exists for all $x$ such in the interval $a < x < b$. Look for points where the derivative might be undefined, such as sharp corners or cusps (e.g., in absolute value functions or root functions like $\sqrt[3]{x^2}$ at $x=0$), or vertical tangents ($f'(x) \to \pm \infty$). Polynomials, rational functions, exponential, and trigonometric functions are differentiable on their domains.
    • Equal Endpoint Values (For Rolle's Theorem ONLY): Calculate the function values at the endpoints, $f(a)$ and $f(b)$, and check if they are equal. This condition is specific to Rolle's Theorem and is not required for the Mean Value Theorem.
  2. Find the Derivative $f'(x)$: Calculate the first derivative of the function $f(x)$ using appropriate differentiation rules.
  3. Find the Value(s) of $c$:
    • For Rolle's Theorem: Set the derivative equal to zero ($f'(c) = 0$) and solve the resulting equation for $c$.
    • For the Mean Value Theorem: Calculate the average rate of change of the function over the interval $[a, b]$ using the formula $m_{sec} = \frac{f(b) - f(a)}{b - a}$. Then, set the derivative at $c$ equal to this average slope ($f'(c) = m_{sec}$) and solve the resulting equation for $c$.
  4. Confirm that $c$ is in the Open Interval $(a, b)$: Check if the value(s) of $c$ obtained in the previous step lie strictly between $a$ and $b$ (i.e., $a < c < b$). Values of $c$ that are equal to $a$ or $b$ or outside the interval $(a, b)$ do not satisfy the conclusion of the theorem.

If all conditions are met and at least one value of $c$ satisfying the conclusion is found within the open interval $(a, b)$, then the theorem is verified for the given function and interval.


Example 1 (Rolle's Theorem Verification). Verify Rolle's Theorem for the function $f(x) = x^2 - 4x + 3$ on the interval $[1, 3]$. Find all values of $c$ that satisfy the conclusion of the theorem.

Answer:

We need to verify the three conditions of Rolle's Theorem for $f(x) = x^2 - 4x + 3$ on the closed interval $[1, 3]$, and then find the value(s) of $c$ in $(1, 3)$ such that $f'(c) = 0$.

Verification of Conditions:

1. Continuity on $[1, 3]$: The function $f(x) = x^2 - 4x + 3$ is a polynomial function. Polynomial functions are continuous for all real numbers. Therefore, $f(x)$ is continuous on the closed interval $[1, 3]$. Condition 1 is satisfied.

2. Differentiability on $(1, 3)$: The derivative of $f(x) = x^2 - 4x + 3$ is $f'(x) = \frac{d}{dx}(x^2) - \frac{d}{dx}(4x) + \frac{d}{dx}(3) = 2x - 4 + 0 = 2x - 4$. This derivative exists for all real numbers. Therefore, $f(x)$ is differentiable on the open interval $(1, 3)$. Condition 2 is satisfied.

3. Equal Endpoint Values ($f(a) = f(b)$): We need to check if $f(1) = f(3)$.

$f(1) = (1)^2 - 4(1) + 3 = 1 - 4 + 3 = 0$.

$f(3) = (3)^2 - 4(3) + 3 = 9 - 12 + 3 = 0$.

Since $f(1) = 0$ and $f(3) = 0$, we have $f(1) = f(3)$. Condition 3 is satisfied.

Since all three conditions of Rolle's Theorem are satisfied for $f(x) = x^2 - 4x + 3$ on $[1, 3]$, the conclusion of the theorem holds, i.e., there exists at least one $c \in (1, 3)$ such that $f'(c) = 0$.

Finding the value(s) of $c$:

We need to find $c \in (1, 3)$ such that $f'(c) = 0$.

We found the derivative $f'(x) = 2x - 4$. Set $f'(c) = 0$:

"$f'(c) = 0$"

"$2c - 4 = 0$"

[Substitute $f'(c)$]

Solve for $c$:

"$2c = 4$"

"$c = 2$"

Confirm $c$ is in $(a, b)$:

The open interval is $(1, 3)$. The value $c=2$ lies strictly between 1 and 3, since $1 < 2 < 3$.

Thus, the value $c=2$ satisfies the conclusion of Rolle's Theorem.

Rolle's Theorem is verified for the given function and interval, and the value of $c$ is 2.


Example 2 (MVT Verification). Verify Lagrange's Mean Value Theorem for the function $f(x) = x^3$ on the interval $[-1, 2]$. Find all values of $c$ that satisfy the conclusion of the theorem.

Answer:

We need to verify the two conditions of the Mean Value Theorem for $f(x) = x^3$ on the closed interval $[a, b] = [-1, 2]$, and then find the value(s) of $c$ in the open interval $(-1, 2)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.

Verification of Conditions:

1. Continuity on $[-1, 2]$: The function $f(x) = x^3$ is a polynomial function. Polynomial functions are continuous for all real numbers. Therefore, $f(x)$ is continuous on the closed interval $[-1, 2]$. Condition 1 is satisfied.

2. Differentiability on $(-1, 2)$: The derivative of $f(x) = x^3$ is $f'(x) = \frac{d}{dx}(x^3) = 3x^2$ (using the Power Rule). This derivative exists for all real numbers. Therefore, $f(x)$ is differentiable on the open interval $(-1, 2)$. Condition 2 is satisfied.

Since both conditions of the Mean Value Theorem are satisfied for $f(x) = x^3$ on $[-1, 2]$, the conclusion of the theorem holds, i.e., there exists at least one $c \in (-1, 2)$ such that $f'(c) = \frac{f(2) - f(-1)}{2 - (-1)}$.

Finding the value(s) of $c$:

First, calculate the average rate of change (the slope of the secant line) over the interval $[a, b] = [-1, 2]$.

$a = -1$, $b = 2$.

$f(a) = f(-1) = (-1)^3 = -1$.

$f(b) = f(2) = (2)^3 = 8$.

Average Rate of Change $= \frac{f(b) - f(a)}{b - a} = \frac{f(2) - f(-1)}{2 - (-1)}$

$= \frac{8 - (-1)}{2 + 1} = \frac{8 + 1}{3} = \frac{9}{3} = 3$

The average rate of change is 3.

Next, find the derivative of the function, which is $f'(x) = 3x^2$.

Now, set $f'(c)$ equal to the average rate of change (3) and solve for $c$:

"$f'(c) = 3$"

"$3c^2 = 3$"

[Substitute $f'(c)=3c^2$]

Divide by 3:

"$c^2 = 1$"

Take the square root:

"$c = \pm \sqrt{1}$"

"$c = 1$ or $c = -1$"

Confirm $c$ is in $(a, b)$:

The open interval specified by the theorem is $(a, b) = (-1, 2)$.

We check which of the values of $c$ lie in this open interval:

  • For $c=1$: Is $-1 < 1 < 2$? Yes, it is. So $c=1$ is in $(-1, 2)$.
  • For $c=-1$: Is $-1 < -1 < 2$? No, it is not. $c=-1$ is an endpoint and is not strictly between $-1$ and $2$.

Thus, only $c=1$ satisfies the conclusion of the theorem.

The Mean Value Theorem is verified for the given function and interval, and the value of $c$ is 1.


Applications of Mean Value Theorems

The Mean Value Theorems, especially Lagrange's MVT, are not just theoretical curiosities; they are fundamental tools used to prove many other significant theorems and to establish properties of functions. While direct application in problem-solving might seem limited to verification exercises, their theoretical power is immense.

In summary, while Mean Value Theorems might appear as exercises in verifying conditions, their true power lies in their role as foundational principles for deriving many other important results in calculus and analysis, and in providing theoretical justification for intuitive concepts about rates of change.