zeros of derivative and increasing function Let $f: \mathbb{R} \to \mathbb{R} $ be a differentiable function. Is it true that $f$ is strictly increasing on $\mathbb R$ if and only if $f'(x) \geq 0$ on $\mathbb R$ and the set $X = \{x \in \mathbb{R} \,| \, f'(x) = 0\}$ is countable?
Suppose $f$ is differentiable and strictly increasing. Then yes, we must have $f'\ge 0$ everywhere. However, we can have $f'=0$ on an uncountable set, in fact a set of positive measure.

Example: Let $U$ be an open set containing the rationals, with $m(U)&lt;1.$ Let $E=\mathbb R\setminus U.$ Then $E$ is closed. Define

$$f(x)=\int_0^x d(t,E)\,dt.$$

Since $d(t,E)$ is continuous, $f'(x) = d(x,E)$ everywhere. We thus have $f'=0$ on $E$ and $f'&gt;0$ on $U.$

Now $m(E)=\infty$ and hence is uncountable. The fact that $f'&gt;0$ on the open dense set $U$ implies $f$ is strictly increasing, and we have our example.
Edit: the op has edited the question to strictly increasing from simply increasing thus my answer is no longer valid. I think this isn't true and here is why, Take $f(x) = -x^2$ when $x&lt;0$, $f(x) = 0$ when $x \in [0,1]$ and $f(x) = x^2-1$ when $x&gt;1$, this function is clearly continuous, differentiable and increasing (not strictly). Now for the set of points between [0,1] f'(x) = 0, but x is an element of the reals so any closed interval such as [0,1] is uncountable.