Monotonic Functions and Stationary Points

Question

Given an arbitrary function $f(x): x \to \mathbb{R}, x \in \mathbb{R}$ and some monotonic function $g(x): x \to \mathbb{R}, x \in \mathbb{R}$, with defined, continuous derivatives, would it be correct to state that

$$ \frac{d}{dx} g(f(x)) \neq 0 \mbox{ } \forall \mbox{ } x \in \mathbb{R} \mbox{ iff } \frac{d}{dx} f(x) \neq 0 \mbox{ } \forall \mbox{ } x \in \mathbb{R} $$

Would it also be correct to state that the stationary and inflection points of $f(x)$ and $g(f(x))$ would coincide?

For example: $e^{(x-2)^2}$. The stationary point of $f(x) = (x-2)^2$ is $x=2$. It would seem to me that, as $\exp(\cdot)$ is a monotonic function, the stationary point would apply to it as well, and so $x=2$ is a stationary point of $g(f(x))$. What I would like to know is whether this is true in general.

Wouldn't that imply that g is not a monotonic function of x? — ck4e, Sep 25 '18 at 20:19
I did not notice the monotonic (1 to 1) issue , maybe cause I was not carefully or the question was edited after my comment, either. @ck4e — dmtri, Sep 26 '18 at 06:27

score 2 · Accepted Answer · answered Sep 25 '18 at 22:15

The one way implication $\,f'(a)=0 \implies (g \circ f)'(a) = 0\,$ holds true by the chain rule. The reverse implication, however, does not hold true in general.

A simple counterexample is $\,f(x)=x\,$ and $\,g(x)=x^3\,$, where $\,g\,$ is strictly increasing and infinitely differentiable. But $\,g'(0)=0\,$ even if $\,f'(0) \ne 0\,$, and $\,f\,$ has neither stationary nor inflection points.

Relevant here is that a strictly increasing function $\,g\,$ can still have its derivative $\,g'\,$ take $\,0\,$ values at some points, even infinitely many points, see Monotone functions and non-vanishing of derivative.

Monotonic Functions and Stationary Points

1 Answers1