The Mean Value Theorem

# The Mean Value Theorem

We are now going to look at a very important theorem in Calculus known as The Mean Value Theorem.

 Theorem 1 (The Mean Value Theorem): If $f$ is a function that satisfies the following conditions: a) $f$ is a continuous function on the closed interval $[a, b]$. b) $f$ is differentiable on the open interval $(a, b)$. Then there must be a value $c \in (a, b)$ where $f'(c)(b - a) = f(b) - f(a)$, that is, there must be a point $(c, f(c))$ whose tangent line has the same slope as the line connecting $(a, f(a))$ and $(b, f(b))$.

From the diagram it should be clear that the slope of the green line is equal to the slope of the pink line. Thus it follows that:

(1)
\begin{align} f'(c) = \frac{f(b) - f(a)}{b - a} \\ f'(c)(b - a) = f(b) - f(a) \end{align}
 Lemma 1: If $f'(x) = 0$ for all $x$ in an interval $[a, b]$, then $f$ had to be a constant on $[a, b]$.
• Proof: First let $x_1$ and $x_2$ be any two values in the interval $[a, b]$ such that $x_1 < x_2$. Since $f$ is continuous on $[a, b]$ we have that $f$ is continuous on $[x_1, x_2]$. Similarly, since $f$ is differentiable on $(a, b)$ we have that $f$ is differentiable on $(x_1, x_2)$. So by the Mean Value Theorem there is a number $c$ in $(x_1, x_2)$ such that:
(2)
$$f(x_2) - f(x_1) = f'(c)(x_2 - x_1)$$
• But by assumption $f'(x) = 0$ on $(a, b)$. So $f'(c) = 0$. Therefore:
(3)
\begin{align} f(x_2) - f(x_1) & = 0(x_2 - x_1) \\ f(x_2) - f(x_1) & = 0 \\ f(x_2) & = f(x_1) \end{align}
• Thus $f$ is a constant on the interval $[a, b]$. $\blacksquare$
 Lemma 2: If $f'(x) > 0$ for all $x$ in an interval $[a, b]$, then $f$ is increasing on $(a, b)$.
• Proof: Let $x_1$, $x_2$ be any two values in the interval $[a, b]$ where $x_1 < x_2$. Since $f$ is continuous on $[a, b]$ we have that $f$ is continuous on $[x_1, x_2]$. Similarly, since $f$ is differentiable on $(a, b)$ we have that $f$ is differentiable on $(x_1, x_2)$. So by the Mean Value Theorem there is a number $c$ in $(x_1, x_2)$ such that:
(4)
$$f(x_2) - f(x_1) = f'(c)(x_2 - x_1)$$
• But by assumption $f'(x) > 0$ on $(a, b)$. So $f'(c) > 0$. Also, since $x_1 < x_2$ we have that $x_2 - x_1 > 0$. Therefore:
(5)
\begin{align} f(x_2) - f(x_1) & > 0 \\ f(x_1) &< f(x_2) \end{align}
• Thus the function $f$ must be increasing on the interval $(a, b)$. $\blacksquare$
 Lemma 3: If $f'(x) < 0$ for all $x$ in an interval $[a, b]$, then $f$ is decreasing on $(a, b)$.
• Proof: Let $x_1$, $x_2$ be values in the closed interval $[a, b]$ such that $x_1 < x_2$. Since $f$ is continuous on $[a, b]$ we have that $f$ is continuous on $[x_1, x_2]$. Similarly, since $f$ is differentiable on $(a, b)$ we have that $f$ is differentiable on $(x_1, x_2)$. So by the Mean Value Theorem there is a number $c$ in $(x_1, x_2)$ such that:
(6)
$$f(x_2) - f(x_1) = f'(c)(x_2 - x_1)$$
• But by assumption $f'(x) < 0$ on $(a, b)$. So $f'(c) < 0$. Also, since $x_1 < x_2$ we have that $x_2 - x_1 > 0$. Therefore:
(7)
\begin{align} f(x_2) - f(x_1) & < 0 \\ f(x_1) & > f(x_2) \end{align}
• Thus the function $f$ is decreasing on the interval $(a, b)$. $\blacksquare$