Mean-Value Theorem

Let $f(x)$ be Differentiable on the Open Interval $(a,b)$ and Continuous on the Closed Interval $[a,b]$. Then there is at least one point $c$ in $(a,b)$ such that

$$f'(c) = {f(b)-f(a)\over b-a}.$$

See also Extended Mean-Value Theorem, Gauss's Mean-Value Theorem

References

Gradshteyn, I. S. and Ryzhik, I. M. Tables of Integrals, Series, and Products, 5th ed. San Diego, CA: Academic Press, pp. 1097-1098, 1993.