Suppose $ f$ is a continuous function on interval $I$ and that the derivative $ f'$ of $f$ exists and is nonnegative everywhere on the interior of $I$. Also suppose that there is no sub-interval of $I$ such that $f'(x)=0$ for all $x$ in the subinterval. Prove that $f$ is a strictly increasing function.
Attempt. I stepped on this when proving that $x-\sin x$ is strictly increasing. One could use the above result, or easier, the MVT on a positive $x>0$ and the nearest $n\pi+\pi/2$ from the left. However i could not transfer this idea to the general case, as stated above, in order to provide a proof of this general result.
Thanks for the help.
Best Answer
$f'(x) \ge 0$ implies that $f$ is (weakly) increasing: $$ a < b \implies f(a) \le f(b) \, , $$ this follows from the mean value theorem.
It remains to exclude that $f(a) = f(b)$ for some $a < b$: Then $$ a \le x \le b \implies f(a) \le f(x) \le f(b) = f(a) \, , $$ i.e. $f$ is constant on $[a, b]$, and therefore $f'(x) = 0$ on $[a, b]$, contradicting the assumption that there is no subinterval on which the derivative is zero.