Proofs of Derivative Applications Facts — Question 3

Question 3

Assume that $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Prove that there exists a number $c\in(a,b)$ such that $$f'(c)=\frac{f(b)-f(a)}{\,b-a\,}.$$

Question 3 - Solution

Define a new function $$g(x)=f(x)-\left(\frac{f(b)-f(a)}{b-a}\right)(x-a).$$

The function $g$ is continuous on $[a,b]$ and differentiable on $(a,b)$ because it is formed from functions with these properties.

Evaluate $g$ at the endpoints: $$g(a)=f(a)-\left(\frac{f(b)-f(a)}{b-a}\right)(a-a)=f(a),$$ $$g(b)=f(b)-\left(\frac{f(b)-f(a)}{b-a}\right)(b-a)=f(b)-(f(b)-f(a))=f(a).$$

Thus, $$g(a)=g(b).$$

By Rolle’s Theorem, there exists a number $c\in(a,b)$ such that $$g'(c)=0.$$

Differentiate $g(x)$: $$g'(x)=f'(x)-\frac{f(b)-f(a)}{b-a}.$$

Substitute $x=c$: $$0=g'(c)=f'(c)-\frac{f(b)-f(a)}{b-a}.$$

Solve for $f'(c)$: $$f'(c)=\frac{f(b)-f(a)}{b-a}.$$

$$\boxed{f'(c)=\dfrac{f(b)-f(a)}{b-a}}$$