 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 Rolle's theorem: $\boxed{f^{\prime }(c)=0}$ for some $c\in (a,b)$.

For the mean value theorem, subtract the endpoint chord:Then $g(a)=g(b)=0$, so Rolle's theorem supplies $c\in (a,b)$ with $g^{\prime }(c)=0$. HenceIf $f^{\prime }(t)>0$ everywhere, applying the mean value theorem to any $x<y$ gives $f(y)-f(x)=f^{\prime }(c)(y-x)>0$. Thus $f$ is strictly increasing. The converse is false: $\boxed{f(t)=t^3}$ is strictly increasing but $f^{\prime }(0)=0$.