Must $g(x)$ be a monotone function?

Question

If $g(x)$ is continuous on $[a,b]$ and it has finite derivative on $(a,b)$ and $g'(x)\not=0$ on $(a,b)$. Is $g(x)$ a monotone function?

Answer 1

Yes, because derivatives, albeit not necessarily being continuous, satisfy the Intermediate Values theorem. Therefore $g'(x)\ne 0$ on $(a,b)$ implies $g'(x) $ has a constant sign.

Answer 2

If you are assuming the derivative exists everywhere, then the derivative indeed takes on a single sign on the interval by Darboux’s theorem, so the function is monotonic.

Answer 3

You can prove it by Intuition, since the derivative of the function is non vanishing in the interval, so there is no point in the interval at which the graph of the function is parallel to the real axis. Hence the graph must be increasing or decreasing, thereby proved to be monotone.

Answer 4

Here is a direct proof which avoids Darboux theorem (intermediate value property of derivatives). Proof below uses letter $f$ instead of $g$ for the function in question.

If $f(a) =f(b) $ then by Rolle's theorem the derivative must vanish somewhere in $(a, b)$. Hence we must have $f(a) \neq f(b) $.

Let us assume that $f(a) <f(b) $ and then we prove that $f$ is strictly increasing on $[a, b] $. Let $x, y$ be such that $a\leq x<y\leq b$. Then Rolle's theorem ensures $f(x) \neq f(y) $. Let us assume $f(x) >f(y) $ and derive a contradiction. We first show that $f(x) \geq f(a) $. If this were not the case then we would have $f(x) <f(a) <f(b) $ and by intermediate value theorem for continuous functions we must have an $x'\in(x, b) $ such that $f(x') =f(a) $ and Rolle's theorem again prevents this. We thus have $f(a) \leq f(x) $. Similarly we can prove that $f(x) < f(b) $. We then have $f(y) <f(x)<f(b) $. And by intermediate value property of continuous functions we must have an $x''\in(y, b) $ with $f(x'') =f(x) $ which sort of contradicts Rolle's theorem.

Therefore $f(x) <f(y) $ and $f$ is strictly increasing in $[a, b] $. If $f(a) >f(b) $ then it can be proved in similar manner that $f$ is strictly decreasing in $[a, b] $.