Rolle's Theorem

maths ✒ rolle

Suppose:

Then $ \exists \phi \in (a,b) $ s.t. $ f'(\phi) = 0 $.

Proof

Since f is continuous and defined over a closed, bounded interval, f is bounded at attains its supremum and infimum.

So let $ f(i) = inf \, f $ and $ f(s) = sup \, f $ where $ i, s \in [a,b] $.

Then if $f(i) = f(s)$:

f must be constant over $[a,b]$, hence $f'(\phi) = 0 ~ \forall \phi \in (a,b)$.

In particular, choose $\phi = \frac {a+b} 2$.

Otherwise, at least one of $i, s \in (a,b)$. WLOG let $s \in (a,b)$ (otherwise consider -f).

Since s is a maximum of f, it must be a local extremum of f therefore by Fermat's Theorem, $f'(s) = 0$ so let $\phi = x$.