I would like to prove the following: "If $f$ is a differentiable function on an open interval and has no local extremum, then $f$ is monotone."
I have seen some questions that asks the same thing in this website, but the answers didn't seem clear to me. Also, since the differentiability condition is given, I would like to make use of it to prove the proposition.
Thank you.
On
It is quite simple. As it is given that the function has no local extremum in the given open interval, $f'\neq 0$ at any point in the interval. So either $f'$ is negative on the entire interval or it is positive on the entire interval. This implies either $f$ is monotonically decreasing or monotonically increasing on the entire interval.
Suppose $f$ is not monotone on the interval $(a,b)$. Since it isn't nondecreasing, there are $c_1,c_2$ in that interval with $c_1 < c_2$ and $f(c_1) > f(c_2)$. By the Mean Value Theorem, there is $c$ in the interval with $$f'(c) = \dfrac{f(c_2)-f(c_1)}{c_2 - c_1} < 0$$ Similarly, since it isn't nonincreasing, there is $d$ in the interval with $f'(d) > 0$.
Case 1: $c < d$.
Since $f$ is continuous on $[c,d]$, it attains a minimum on the interval $[c,d]$. That minimum can't be at $c$ because $f'(c) < 0$, and can't be at $d$ because $f'(d) > 0$, so it must be at some point $e$ with $c < e < d$. Then $e$ is a local maximum of $f$.
Case 2: $d < c$.
By a similar argument, the maximum of $f$ on the interval $[d,c]$ must be at some $e$ with $d < e < c$, and this is a local maximum of $f$.