= Solution
The <mean value theorem> states that a real-valued function continuous on $[a,b]$ and differentiable on $(a,b)$, with $a<b$, has a point $\xi\in(a,b)$ such that
$$
\boxed{f'(\xi)=\frac{f(b)-f(a)}{b-a}.}
$$
For the proof, subtract the secant line and put $h(t)=f(t)-f(a)-(t-a)[f(b)-f(a)]/(b-a)$. Then $h(a)=h(b)=0$. If $h$ is identically zero, every interior point works. Otherwise it has a positive maximum or a negative minimum. The <extreme value theorem> supplies that extremum, and its nonzero value makes its location interior. At an interior differentiable extremum, the difference quotients on the two sides have opposite weak signs, so their common <limit> is zero. Hence $h'(\xi)=0$, proving the formula. This also gives the <Rolle theorem> argument rather than merely citing it.
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