real analysis - If $f$ is nowhere differentiable does it follow that $f$ is monotonic at no point?

Let $f \colon \mathbb{R} \to \mathbb{R}$ be a continuous functions that is nowhere differentiable. From this question (Does there exist a nowhere differentiable, everywhere continous, monotone somewhere function?) , I know that it follows
that $f$ is monotone on no interval.

Let $x$ be a real number. We say that $f$ is non-decreasing at $x$ if there is a neighborhood of $x$, $N_x$, such that $\frac{f(y)-f(x)}{y-x} \ge 0$ if $y \in N_x-\{x\}$.

If a function is continuous everywhere and differentiable nowhere,
does it follow that it is monotonic at no point? If this is not the
case can you please give a counterexample?

Answer

This does not follow: here's a cheap way to modify a given $f$ so that it becomes monotone at a point. Consider a local minimum $x_0$ of $f$. Let's assume that $x_0=f(x_0)=0$. Then $f(x)\ge 0$ in a neighborhood of $0$, so
$$g(x) = \begin{cases} f(x) & x\ge 0\\ -f(x) & x<0 \end{cases}$$
is monotone at $0$. If I'm spectacularly unlucky here, then $g$ is now differentiable at $0$, but then I can simply redefine it as $2f(x)$ for $x\ge 0$.