Answer
ASSUME that $f(x)$ is not constant on $(a, b)$ and
using the Mean Value Theorem, arrive at a contradiction.
(Please see details in step-by-step)
Work Step by Step
ASSUME that $f(x)$ is not constant on $(a, b)$ .
there exists $x_{1}$ and $x_{2}$ in $(a, b)$ such that $f(x_{1})\neq f(x_{2})$ .
By the Mean Value Theorem, there exists $c$ in $(a, b)$
such that $f^{\prime}(c)=\displaystyle \frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}$
$f^{\prime}(c)\neq 0$ because the numerator is not zero.
But, we were given: $f^{\prime}(x)=0$ for all $x$ in $(a, b)$ .
We have a contradiction, our ASSUMPTION was wrong.
Therefore, $f(x)$ is constant on $(a, b)$ .
