Proofs of Derivative Applications Facts — Question 9

Question 9 - Solution

Let $x_1,x_2\in[a,b]$ with $x_1<x_2$. We will show that $$f(x_1)=f(x_2).$$

Since $f$ is continuous on $[x_1,x_2]$ and differentiable on $(x_1,x_2)$, the Mean Value Theorem applies. Thus, there exists a number $c\in(x_1,x_2)$ such that $$f'(c)=\frac{f(x_2)-f(x_1)}{x_2-x_1}.$$

By assumption, $f'(c)=0$, so $$0=\frac{f(x_2)-f(x_1)}{x_2-x_1}.$$

Since $x_2-x_1\neq 0$, it follows that $$f(x_2)-f(x_1)=0.$$

Therefore, $$f(x_2)=f(x_1).$$

Because $x_1$ and $x_2$ were arbitrary points in $[a,b]$, we conclude that $f(x)$ has the same value at every point in the interval.

$$\boxed{f\text{ is constant on }[a,b]}$$