Finding inflection points of the function $\sqrt[3]{x}$

calculusderivatives

I have been working on a problem where I have to find critical number and inflection point for the function $\sqrt[3]{x}$

While I have been able to find critical number, I'm not sure how to find the inflection point for the function as for this particular function I cannot assign double derivative to be zero and then solve for x. The double derivative for other points indicates that the inflection point is between -1 and 1, but I'm not able to find any more ideas on how to approach this.

$f''(-1) = 2/9 > 0$

$f''(1) = -2/9 < 0$

Best Answer

The inflection point is $(0,0)$. To see this, note that $f(x) = \sqrt[3]{x}$ is twice differentiable on $\mathbb R^* =\mathbb R \setminus \{0\}$:

$$f'(x) = \frac{1}{3\sqrt[3]{x^2}}, f''(x) = - \frac{2}{9\sqrt[3]{x^5}} .$$

$f$ is not differentiable at $x = 0$ (but it has the improper derivative $+\infty$ at $0$, see for example Having trouble understanding condition for Rolle's Theorem (russian translation)).

You can now apply 1.3 Theorem 4.4.3 to see that $f$ is concave up on $(-\infty,0]$ and concave down on $[0,\infty)$. According to 1.4 Definition 4.4.4 the point $(0,0)$ is the only inflection point of $f$.

Intuitively this is clear if we look at $f''$: We have $\lim_{x \to 0^+} f''(x) = -\infty$ and $\lim_{x \to 0^-} f''(x) = +\infty$, i.e. $f''$ has an "improper change of sign" at $0$.

Update:

You can also easily prove the following theorem:

Let $f : I \to J$ be a bijection between to real intervals. Then $(a,f(a))$ is an inflection point of $f$ if and only $(f(a),a)$ is an inflection point of $f^{-1} : J \to I$.

This true because the graph of $f^{-1}$ is obtained from ther graph of $f$ by a reflection at the diagonal line $y = x$.

In your example $f^{-1}(x) = x^3$ which has exactly one inflection point at $(0,0)$.

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