# proof of mean value theorem

Define $h(x)$ on $[a,b]$ by

 $h(x)=f(x)-f(a)-\left(\frac{f(b)-f(a)}{b-a}\right)(x-a)$

Clearly, $h$ is continuous on $[a,b]$, differentiable on $(a,b)$, and

 $\begin{array}[]{ccl}h(a)&=&f(a)-f(a)=0\\ h(b)&=&f(b)-f(a)-\left(\frac{f(b)-f(a)}{b-a}\right)(b-a)=0\\ \end{array}$

Notice that $h$ satisfies the conditions of Rolle’s Theorem. Therefore, by Rolle’s Theorem there exists $c\in(a,b)$ such that $h^{\prime}(c)=0$.
However, from the definition of $h$ we obtain by differentiation that

 $h^{\prime}(x)=f^{\prime}(x)-\frac{f(b)-f(a)}{b-a}$

Since $h^{\prime}(c)=0$, we therefore have

 $f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$

as required.

## References

• 1 Michael Spivak, Calculus, 3rd ed., Publish or Perish Inc., 1994.
Title proof of mean value theorem ProofOfMeanValueTheorem 2013-03-22 12:40:57 2013-03-22 12:40:57 Andrea Ambrosio (7332) Andrea Ambrosio (7332) 5 Andrea Ambrosio (7332) Proof msc 26A06