3.2: The Mean Value Theorem


We will discuss the Mean Value Theorem in this section.

Rolle's Theorem


Rolle's Theorem is a simpler case of the Mean Value Theorem.

Rolle's Theorem: Let $f$ be a function that satisfies the following hypotheses:
  1. $f$ is continuous on the closed interval $[a,b]$
  2. $f$ is differentiable on the open interval $(a,b)$
  3. $f(a) = f(b)$
Then there is a number $c\in (a, b)$ such that $f'(c) = 0$.

Suppose $s = f(t)$ is a position function. If the object is in same place at two different time points $t = a$ and $t = b$, then $f(a) = f(b)$.

Rolle's Theorem says there must exist $t = c$ between $a$ and $b$ where $f'(c) = 0$, or in other words, the velocity is zero.

You can see this in action by tossing a ball upwards; the ball's upward velocity reaches zero (which coincides with a local maximum!).

Mean Value Theorem


Imagine if the heights $f(a)$ and $f(b)$ were different. This is called the Mean Value Theorem.

Mean Value Theorem: Let $f$ be a function that satisfies the following hypotheses:
  1. $f$ is continuous on the closed interval $[a,b]$
  2. $f$ is differentiable on the open interval $(a,b)$
Then there is a number $c\in (a, b)$ such that \[f'(c) = \dfrac{f(b) - f(a)}{b - a}\] or multiplying both sides by $(b - a)$: \[f(b) - f(a) = f'(c)(b - a)\]
Verify that the function \[f(x) = x^3 - x, \ \ [0, 4]\] satisfies the hypotheses of the MVT. Then find all numbers $c$ that satisfy the conclusion of the MVT.

Consequences of the MVT


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

Let $x_1, x_2 \in (a, b)$ where $x_1 < x_2$.

Because $f$ is differentiable on $(a, b)$, it must be differentiable on $(x_1, x_2)$ and continuous on $[x_1, x_2]$. Therefore the hypotheses of MVT are satisfied.

Applying MVT on the interval $[x_1, x_2]$ gives us a number $c$ where $x_1 < c < x_2$ and \[f(x_2) - f(x_1) = f'(c)(x_2 - x_1)\]

Because $f'(x) = 0$, in particular $f'(c) = 0$, the equation above becomes \[f(x_2) - f(x_1) = 0\] or $f(x_2) = f(x_1)$.

This means all the heights $f(x)$ are the same in $(a, b)$. Therefore $f(x) = c$, or what we call $f$ is constant.

This is saying if you take the derivative of $f(x)$ and it's equal to zero, then $f(x) = c$ for some $c \in \mathbb{R}$.

The next consequence is the one we are after: the link to the antiderivatives.

Corollary: If $f'(x) = g'(x)$ for all $ x \in (a, b)$, then $f - g$ is constant on $(a, b)$, that is, \[f(x) = g(x) + c\] where $c$ is a constant.

Let $F(x) = f(x) - g(x)$. Then \[F'(x) = f'(x) - g'(x) = 0\] because $f'(x) = g'(x)$. By the previous theorem $F$ is constant; that is, $f - g$ is constant.