Rolle's Theorem (Baby MVT)

Rolle's Theorem

Let $f : [a, b] \to R$ be continuous on $[a, b]$ be differentiable at least on $(a, b)$ . If $f (a) = f (b)$ then $\exists c \in (a, b)$ such that $f^{'} (c) = 0$ .

All this comes down to is applying the Interior Extremum Theorem again.

Proof

Let $f$ be continuous on $[a, b]$ . Then by the Extreme Value Theorem then $f$ has an absolute maximum and an absolute minimum. Now if either of those are in-between $c \in (a, b)$ then apply the Interior Extremum Theorem then $f^{'} (c) = 0$ .

Now what if $c = a, b$ ? If that's the case then $f$ is constant (because the max and the min are at the same point $c$ , so the function cannot increase nor decrease at all) on $[a, b]$ which has a zero-derivative everywhere (just pick one).

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