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Mean Value Theorem

The mean value theorem says that somewhere inside an interval, a function's instantaneous rate of change equals its average rate of change. Geometrically, a tangent to the curve is parallel to the secant joining its endpoints. It extends Rolle's theorem to functions whose endpoint values need not be equal.
MEAN VALUE THEOREM
Let a<ba<b, and let f:[a,b]→Rf:[a,b]\to\mathbb{R} be continuous on [a,b][a,b] and differentiable on (a,b)(a,b). Then there is at least one point c∈(a,b)c\in(a,b) such that
f(b)−f(a)=f′(c)(b−a).\begin{equation*}f(b)-f(a)=f'(c)(b-a).\end{equation*}
Equivalently,
f′(c)=f(b)−f(a)b−a.\begin{equation*}f'(c)=\frac{f(b)-f(a)}{b-a}.\end{equation*}

Proof

Define
h(x)=(b−a)f(x)−x[f(b)−f(a)].\begin{equation*}h(x)=(b-a)f(x)-x\bigl[f(b)-f(a)\bigr].\end{equation*}
At the endpoints,
h(a)=h(b)=bf(a)−af(b).\begin{equation*}h(a)=h(b)=bf(a)-af(b).\end{equation*}
Also, hh is continuous on [a,b][a,b] and differentiable on (a,b)(a,b). By Rolle's theorem, there is a point c∈(a,b)c\in(a,b) for which h′(c)=0h'(c)=0. Since
h′(x)=(b−a)f′(x)−[f(b)−f(a)],\begin{equation*}h'(x)=(b-a)f'(x)-\bigl[f(b)-f(a)\bigr],\end{equation*}
substituting x=cx=c gives
0=(b−a)f′(c)−[f(b)−f(a)].\begin{equation*}0=(b-a)f'(c)-\bigl[f(b)-f(a)\bigr].\end{equation*}
Rearranging proves
f(b)−f(a)=f′(c)(b−a).\begin{equation*}f(b)-f(a)=f'(c)(b-a).\end{equation*}

References

  • [Apostol1967]Apostol, Tom M. Calculus, Volume I: One-Variable Calculus, with an Introduction to Linear Algebra. Second edition. John Wiley \& Sons, 1967.