= Solution
Suppose $f$ is continuous on $[a,b]$, differentiable on $(a,b)$ and satisfies $f(a)=f(b)$, with $a<b$. Apply the <extreme value theorem> to both $f$ and $-f$ to obtain a maximum and a minimum. If $f$ is constant, every interior derivative is zero. Otherwise some value differs from the common endpoint value, so either a maximum or a minimum occurs at an interior point $c$. At an interior maximum, difference quotients are nonpositive from the right and nonnegative from the left; since the <derivative> exists, it must be zero. The minimum case is identical with the inequalities reversed. This proves \b[Rolle's theorem:] $\boxed{f'(c)=0}$ for some $c\in(a,b)$.
For the <mean value theorem>, subtract the endpoint chord:
$$
g(t)=f(t)-f(a)-\frac{f(b)-f(a)}{b-a}(t-a).
$$
Then $g(a)=g(b)=0$, so <Rolle's theorem> supplies $c\in(a,b)$ with $g'(c)=0$. Hence
$$
\boxed{f'(c)=\frac{f(b)-f(a)}{b-a}.}
$$
If $f'(t)>0$ everywhere, applying the <mean value theorem> to any $x<y$ gives $f(y)-f(x)=f'(c)(y-x)>0$. Thus $f$ is <strictly increasing>. \b[The converse is false:] $\boxed{f(t)=t^3}$ is <strictly increasing> but $f'(0)=0$.
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