Mean Value Theorem

Introduction to the Mean Value Theorem

The mean value theorem is a fundamental result in calculus that links the average rate of change of a function over an interval to the instantaneous rate of change at some point within that interval.

Conditions for the Mean Value Theorem

To apply the mean value theorem, the following conditions must be satisfied:

The function $f$ must be continuous over the closed interval $[a, b]$ .
The function $f$ must be differentiable over the open interval $(a, b)$ .

Exam Tip

Always ensure to justify that the function satisfies both conditions of continuity and differentiability before applying the theorem in an exam.

Understanding the Theorem

If the above conditions are met, then there exists at least one point $c$ in the open interval $(a, b)$ such that:

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

This means that there exists a point within the interval where the instantaneous rate of change (the derivative) is equal to the average rate of change over the interval.

Key Concept

The mean value theorem guarantees the existence of such a point but does not specify where it is.

Worked Example

A social sciences researcher is using a function

m(t)

to model the total mass of all the garden gnomes appearing on lawns in a particular neighborhood over time

t

(measured in days). The function

m(t)

is twice-differentiable. The table below provides selected values of

m(t)

over the interval

0 \le t \le 12

.

$t$ (days)	0	3	7	10	12
$m(t)$ (kg)	24.9	36.0	70.3	89.7	89.1

Justify why there must be at least one time $t$ within $10 \le t \le 12$ at which $m'(t) = -0.3$ kilograms per day.

Solution:

Verify Continuity and Differentiability:
- Since $m(t)$ is twice-differentiable, it is also differentiable.
- A differentiable function is automatically continuous.
- Thus, $m(t)$ is continuous and differentiable over the interval $[10, 12]$ .
Calculate the Average Rate of Change: $\text{Average rate of change} = \frac{m(12) - m(10)}{12 - 10} = \frac{89.1 - 89.7}{2} = \frac{-0.6}{2} = -0.3 \text{ kg/day}$
Apply the Mean Value Theorem:
- Since $m(t)$ is continuous on $[10, 12]$ and differentiable on $(10, 12]$ , by the mean value theorem, there exists at least one point $t$ in $(10, 12)$ such that: $m'(t) = -0.3 \text{ kg/day}$

Thus, the mean value theorem guarantees the existence of such a point within the given interval.

Practice Questions

Practice Question

Consider the function $f(x) = x^3 - 3x + 2$ . Verify that the function satisfies the conditions of the mean value theorem over the interval $[-2, 2]$ and find the value of $c$ where the theorem applies.
Let $g(x) = \sin(x)$ over the interval $[0, \pi]$ . Show that there is a point $c$ in $(0, \pi)$ where $g'(c) = \frac{g(\pi) - g(0)}{\pi - 0}$ .

Rolle's Theorem

Introduction to Rolle's Theorem

Rolle's theorem is a special case of the mean value theorem. It applies when the function values at the endpoints of the interval are equal.

Conditions for Rolle's Theorem

The function $f$ must be continuous over the closed interval $[a, b]$ .
The function $f$ must be differentiable over the open interval $(a, b)$ .
The function values at the endpoints must be equal, i.e., $f(a) = f(b)$ .

Statement of the Theorem

If the above conditions are met, then there exists at least one point $c$ in the open interval $(a, b)$ such that:

$f'(c) = 0$

This means that there is a point within the interval where the derivative (instantaneous rate of change) is zero, indicating a horizontal tangent line, which corresponds to a local maximum or minimum.

Key Concept

Rolle's theorem guarantees the existence of at least one point where the derivative is zero within the interval.

Consider the function

h(x) = x^2 - 4x + 4

over the interval

[0, 4]

.

Verify the Conditions:
- $h(x)$ is a polynomial, so it is continuous and differentiable everywhere.
- $h(0) = 0^2 - 4 \cdot 0 + 4 = 4$
- $h(4) = 4^2 - 4 \cdot 4 + 4 = 4$
- Hence, $h(0) = h(4)$ .
Apply Rolle's Theorem:
- Since the conditions are satisfied, there must be at least one point $c$ in $(0, 4)$ where $h'(c) = 0$ .
- Compute the derivative: $h'(x) = 2x - 4$ .
- Set the derivative to zero: $2c - 4 = 0 \implies c = 2$ .

Thus, $h'(2) = 0$ and $c = 2$ is the point where the derivative is zero within the interval.

Summary and Key Takeaways

The mean value theorem connects the average rate of change of a function over an interval to the instantaneous rate of change at some point within that interval.
To apply the mean value theorem, ensure the function is continuous on the closed interval and differentiable on the open interval.
Rolle's theorem is a special case of the mean value theorem, applicable when the function values at the endpoints are equal.
Both theorems guarantee the existence of specific points but do not pinpoint their exact locations.

Glossary

Continuous Function: A function without any breaks, jumps, or holes in its domain.
Differentiable Function: A function that has a derivative at every point in its domain.
Instantaneous Rate of Change: The derivative of a function at a specific point.
Average Rate of Change: The change in the value of a function over an interval divided by the length of the interval.

Practice Questions Solutions

Question 1:
- The function $f(x) = x^3 - 3x + 2$ is continuous and differentiable everywhere (being a polynomial).
- The average rate of change over $[-2, 2]$ is: $\frac{f(2) - f(-2)}{2 - (-2)} = \frac{(2^3 - 3 \cdot 2 + 2) - ((-2)^3 - 3 \cdot (-2) + 2)}{4} = \frac{(8 - 6 + 2) - (-8 + 6 + 2)}{4} = \frac{4 - 0}{4} = 1$
- By the mean value theorem, there exists $c \in (-2, 2)$ such that: $f'(c) = 1$
- Compute the derivative: $f'(x) = 3x^2 - 3$ .
- Solve $3c^2 - 3 = 1$ : $3c^2 = 4 \implies c^2 = \frac{4}{3} \implies c = \pm \sqrt{\frac{4}{3}} = \pm \frac{2}{\sqrt{3}}$
Question 2:
- The function $g(x) = \sin(x)$ is continuous and differentiable on $[0, \pi]$ .
- The average rate of change over $[0, \pi]$ is: $\frac{g(\pi) - g(0)}{\pi - 0} = \frac{\sin(\pi) - \sin(0)}{\pi} = \frac{0 - 0}{\pi} = 0$
- By the mean value theorem, there exists $c \in (0, \pi)$ such that: $g'(c) = 0$
- Compute the derivative: $g'(x) = \cos(x)$ .
- Solve $\cos(c) = 0$ : $c = \frac{\pi}{2}$

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Graphs of Functions & Their Derivatives

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