= Solution
The <mean value theorem> states that a function <continuous> on $[a,b]$ and <differentiable> on $(a,b)$, where $a<b$, has a point $\xi\in(a,b)$ such that
$$
\boxed{f'(\xi)=\frac{f(b)-f(a)}{b-a}.}
$$
To prove it using <Rolle's theorem>, subtract the secant line:
$$
h(t)=f(t)-f(a)-\frac{f(b)-f(a)}{b-a}(t-a).
$$
This function has the required <continuity> and <differentiability>, with $h(a)=h(b)=0$. <Rolle's theorem> gives $h'(\xi)=0$, which is exactly the claimed equation. Thus a tangent slope somewhere in the interval equals the overall secant slope.
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