Introduction

This topic introduces the Mean Value Theorem, one of the most important existence theorems in calculus. The essential knowledge tells us that if a function f is continuous over the interval $[a,b]$ and differentiable over the interval $(a,b)$, then the Mean Value Theorem guarantees a point within that open interval where the instantaneous rate of change equals the average rate of change over the interval. This theorem allows us to draw powerful conclusions about function behavior without finding exact values.

Understanding the Mean Value Theorem

The Theorem Statement

Mean Value Theorem (MVT): If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists at least one value $c$ in $(a,b)$ such that:

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

Interpreting the Theorem

Left Side: $f^{\prime }(c)$ is the instantaneous rate of change at $x=c$

Right Side: $\frac{f(b)-f(a)}{b-a}$ is the average rate of change over $[a,b]$

Geometric Meaning: There exists a point where the tangent line is parallel to the secant line connecting the endpoints.

Graphical Representation of MVT 

Credits to Paul's Math Notes.

The above graph shows a section of a graph over interval $[A,B]$ with a point $C$ where the tangent line follows the same slope as the secant line over the interval $[A,B]$ showing an instance of the MVT where the derivative (the tangent line) has a slope equivalent to that of the average ( secant line)

Example 1:

For $f(x)=x^2$ on $[1,3]$, the average rate of change is:

$\frac{f(3)-f(1)}{3-1}=\frac{9-1}{2}=4$

Find where $f^{\prime }(c)=4$: $f^{\prime }(x)=2x$ $2c=4$ $c=2$

At $x=2$, the tangent line has slope 4, matching the secant line slope.

Conditions for the Mean Value Theorem

Required Conditions

The MVT requires TWO conditions:

Condition 1: $f$ is continuous on $[a,b]$ (closed interval)

Condition 2: $f$ is differentiable on $(a,b)$ (open interval)

If either condition fails, the MVT does not apply.

Example 2:

For $f(x)=|x|$ on $[-1,1]$:

The function is continuous on $[-1,1]$ but not differentiable at $x=0$.

Average rate of change: $\frac{1-1}{2}=0$

The derivative is never $0$ on the interval (it's $-1$ or $1$).

The MVT does not apply because differentiability fails.

Finding the Value of c

Solving for c

When the MVT applies, solve $f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$ to find specific values.

Example 3:

Find $c$ guaranteed by the MVT for $f(x)=x^3-3x$ on $[0,2]$.

Check conditions: polynomial is continuous and differentiable ✓

Average rate of change: $\frac{f(2)-f(0)}{2-0}=\frac{2-0}{2}=1$

Set $f^{\prime }(c)=1$:

$f^{\prime }(x)=3x^2-3$

$3c^2-3=1$

$3c^2=4$

$c=\frac{2\sqrt{3}}{3}\approx 1.155$

Applications and Justifications

Speed Application

Example 4:

A car travels 100 miles in 2 hours. Prove that at some moment, the car was traveling exactly 50 mph.

Average velocity: $\frac{100}{2}=50$ mph

Assuming continuous position and velocity exists, the MVT guarantees that at some time $c$ in the 2-hour interval: $s^{\prime }(c)=50$ mph

The instantaneous velocity equaled the average velocity.

Writing Complete Justifications

A complete MVT justification includes:

Example 5:

Justify that $f(x)=x^2+1$ satisfies the MVT on $[1,3]$.

$f(x)=x^2+1$ is a polynomial, so it is continuous on $[1,3]$ and differentiable on $(1,3)$.

By the Mean Value Theorem, there exists $c$ in $(1,3)$ such that: $f^{\prime }(c)=\frac{f(3)-f(1)}{3-1}=\frac{10-2}{2}=4$

Practice Section