October 6th, 2026

Updated 7 Oct 2026Notes (PDF)

[0.0.44]Theorem(Rolle's Theorem)#

Suppose $f: [a, b] \to \bR$ is a continuous function that is differentiable on $(a, b)$ . If $f(a) = f(b)$ , then there is some $c \in (a, b)$ such that $f'(c) = 0$ .

Proof.

By the Extreme Value Theorem, $f$ attains a maximum and minimum. If both extrema occur at the endpoints, then $f$ is a constant function, and so $f'(c) = 0$ for all $c \in (a, b)$ . Otherwise, note that $f$ attains a local extremum at some $c \in (a, b)$ . By the Stationary Point Theorem, note $f'(c) = 0$ .

[0.0.45]Theorem(Mean Value Theorem)#

Suppose $f: [a, b] \to \bR$ is a continuous function that is differentiable on $(a, b)$ . Then there is some $c \in (a, b)$ such that

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

Proof.

Let us consider a graph for intuition.

We seek to show the distance between the curve $f(x)$ and the secant line from $a$ to $b$ has derivative $0$ at some point. Note the secant line, by point-slope, can be written as

y - f(a) = \left(\frac{f(b) - f(a)}{b - a}\right)(x - a).

Then the distance function is given by

d(x) = f(x) - \left[f(a) + \left(\frac{f(b) - f(a)}{b - a}\right)(x - a) \right].

It is quick to show that $d(a) = d(b)$ . Moreover, $d$ is differentiable on $(a, b)$ . By Mean Value Theorem, we have that there is some $c \in (a, b)$ such that $d'(c) = 0$ . Rearrangement gives us

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

completing the proof.