The maximum number of stationary points a cube root function can have

algebra-precalculuscalculusfunctionsinverse function

When I came across this question, I knew that a cube root function in the form:

$$y=a(b(x-h))^{\frac{1}{3}}-k$$

For cubics, they can have one stationary point at $x=h$.
But I also know that a cube root function is the inverse of a cubic function, hence it can also have $0$ or $2$ stationary points right? Is this a correct assumption to make?

Best Answer

$g(x) = a(b(x-h))^{1/3}-k = a b^{1/3} \cdot (x-h)^{1/3}-k$ has no stationary points at all. $g$ is not differentiable at $x=h$, and at all other points is $g'(x)$ positive or negative, depending on the sign of $a b^{1/3}$.

More generally, the inverse function of a differentiable function never has stationary points:

  • Let $I \subset \Bbb R$ be a (finite or infinite) interval,
  • $f: I \to \Bbb R$ a differentiable and injective function,
  • $g: f(I) \to I$ the inverse function.

Then $f(g(y)) = y$ for all $y \in f(I)$, and if $g$ is differentiable at $y_0$ then $$ f'(g(y_0)) \cdot g'(y_0) = 1 $$ so that necessarily $g'(y_0) \ne 0$.

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