$f'(x)\neq0 \Rightarrow$ f is monotonic

Question

let $\ f:[a,b] \rightarrow \Bbb{R} $, given that $f'(x) \neq 0$ can we conclude that $f$ is monotonic ?

Answer 1

Yes, assuming $f$ is differentiable for all $x \in [a,b]$ including the endpoints.

Actually, we can say it's strictly monotonic. There's two steps to proving this:

$(1)$ Note that if $f:[a,b] \to \mathbb{R}$ is continuous, then $f$ is injective if and only if it's strictly monotonic. Proof: one direction is trivial, the other follows from the IVT. (in our case, we know $f$ is differentiable, so it is of course continuous).

$(2)$ If $f:[a,b] \to \mathbb{R}$ is differentiable and $f'(x) \neq 0$ for all $x \in [a,b]$, then $f$ is injective. Proof: suppose otherwise. Then there are $x,y \in [a,b]$ such that $\frac{f(x) - f(y)}{x-y} = 0$. We derive a contradiction from the mean value theorem.

The comments are all incorrect and misleading. The continuity of the derivative is not needed here.