The mean value theorem connects the derivative at one point to the behavior of a function across an interval. It is the reason a positive derivative means increasing, and it is behind every error estimate in calculus. You met it in Calculus 1; here is the proof, which rests on the extreme value theorem and differentiability.
Key idea
Rolle's theorem. If $f$ is continuous on $[a, b]$, differentiable on $(a, b)$, and $f(a) = f(b)$, then $f'(c) = 0$ for some $c \in (a, b)$. Mean value theorem. If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then $f'(c) = \frac{f(b) - f(a)}{b - a}$ for some $c \in (a, b)$.
Why it works
Rolle. By the EVT, $f$ has a maximum and a minimum on $[a, b]$. If both occur at the ends, $f$ is constant (since $f(a) = f(b)$) and any $c$ works. Otherwise one of them is at an interior point $c$. At an interior maximum, the difference quotient is $\le 0$ for $h > 0$ and $\ge 0$ for $h < 0$, so the derivative, the common limit, is $0$. MVT. Subtract the secant line: $g(x) = f(x) - \frac{f(b) - f(a)}{b - a}(x - a)$ has $g(a) = g(b) = f(a)$. Rolle gives $g'(c) = 0$, which is $f'(c) = \frac{f(b) - f(a)}{b - a}$. $\blacksquare$
Worked examples
Worked example: A quadratic always gives the midpoint
Find $c$ for $f(x) = 3x^{2} - 2x$ on $[0, 4]$.
- Secant slope: $\frac{f(4) - f(0)}{4} = \frac{40}{4} = 10$.
- $f'(c) = 6c - 2 = 10$ gives $c = 2$.
Answer: $c = 2$, the midpoint. (For any quadratic, $f'$ is linear and the MVT point is the midpoint.)
Worked example: A cubic with one admissible $c$
Find all $c$ for $f(x) = x^{3} - 3x$ on $[-1, 2]$.
- Secant slope: $\frac{f(2) - f(-1)}{3} = \frac{2 - 2}{3} = 0$.
- $f'(c) = 3c^{2} - 3 = 0$ gives $c = \pm 1$.
Answer: Only $c = 1$ is in the open interval $(-1, 2)$.
Worked example: Counting roots with Rolle
Show that $x^{5} + x - 3 = 0$ has exactly one real root.
- It has at least one: the polynomial is negative at $0$ and positive at $2$ (IVT).
- If it had two, Rolle would give a zero of the derivative $5x^{4} + 1$ between them, but $5x^{4} + 1 > 0$.
Answer: Exactly one.
Common mistake
Skipping the hypotheses, or the open interval. $|x|$ on $[-1, 1]$ has secant slope $0$ but no point with derivative $0$: it is not differentiable at $0$. Likewise $x^{2/3}$ on $[-1, 1]$. And the theorem's $c$ lies strictly inside $(a, b)$: in the cubic example, $c = -1$ solves the equation but is an endpoint, so it is not the theorem's point.