Calculus I · The Mean Value Theorem

First, the easy case: Rolle’s theorem

Start with a picture so simple it feels like it can’t be saying much. Suppose a smooth curve starts and ends at the same height — $f(a) = f(b)$. Draw any such curve in the air with your finger. To come back down to where you started, you had to stop climbing at some point and turn around: there is a peak, or a valley, somewhere strictly between the ends. And at that turning point, the curve is momentarily flat — its slope is zero.

That is the entire content of Rolle’s theorem.

Rolle’s Theorem. If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a) = f(b)$, then there is at least one point $c$ in $(a,b)$ where $f'(c) = 0$.

The proof rests on one small, sharp fact about what the derivative must do at a peak — a fact worth isolating, because we will reuse it when we hunt for maxima in Chapter 8.

◆ Fermat’s lemma: at an interior peak, the slope is zero

Claim. If $f$ has a local maximum at an interior point $c$ (and $f'(c)$ exists), then $f'(c) = 0$. (Same for a local minimum.)

Why. Look at the difference quotient $\dfrac{f(c+h) - f(c)}{h}$ near the peak. Because $c$ is a maximum, the numerator $f(c+h) - f(c)$ is $\le 0$ for every small $h$ — you can’t go higher than the top. Now watch the sign of the whole quotient from each side: - From the right ($h > 0$): numerator $\le 0$, denominator $> 0$, so the quotient is $\le 0$. Taking $h\to0^+$ gives $f'(c) \le 0$. - From the left ($h < 0$): numerator $\le 0$, denominator $< 0$, so the quotient is $\ge 0$. Taking $h\to0^-$ gives $f'(c) \ge 0$.

The derivative exists, so both one-sided limits equal $f'(c)$. The only number that is both $\le 0$ and $\ge 0$ is $0$. So $f'(c) = 0$. $\blacksquare$

With that in hand, Rolle falls out cleanly:

◆ Proof of Rolle’s theorem

By the Extreme Value Theorem, $f$ attains a maximum and a minimum somewhere on $[a,b]$. Two cases. - If both the max and the min happen at the endpoints, then since $f(a) = f(b)$, the highest and lowest values of $f$ are equal — so $f$ is constant, and $f'(c) = 0$ at every interior $c$. - Otherwise, at least one of them occurs at an interior point $c$. By Fermat’s lemma, $f'(c) = 0$ there.

Either way, some interior $c$ has $f'(c) = 0$. $\blacksquare$

(Figure 6.1: a curve with $f(a) = f(b)$, its interior peak marked, and a flat tangent line drawn at that peak.)

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