# What are the points of inflection, if any, of f(x)=x^(1/3) ?

Dec 7, 2015

This is an interesting example. The point $\left(x , f \left(x\right)\right) = \left(0 , f \left(0\right)\right) = \left(0 , 0\right)$ can be considered to be a point of inflection because the graph of $f$ changes from concave up to concave down as $x$ increases through zero. However, $f ' \left(0\right)$ and $f ' ' \left(0\right)$ do not exist.

#### Explanation:

For $f \left(x\right) = {x}^{\frac{1}{3}}$, we have $f ' \left(x\right) = \frac{1}{3} {x}^{- \frac{2}{3}} = \frac{1}{3 {x}^{\frac{2}{3}}}$ when $x \ne 0$ and $f ' ' \left(x\right) = - \frac{2}{9} {x}^{- \frac{5}{3}} = - \frac{2}{9 {x}^{\frac{5}{3}}}$ when $x \ne 0$. Therefore, $f ' ' \left(x\right) > 0$ when $x < 0$ and $f ' ' \left(x\right) < 0$ when $x > 0$.

However, $f ' \left(0\right)$ does not exist since ${\lim}_{h \to 0} \frac{f \left(0 + h\right) - f \left(0\right)}{h} = {\lim}_{h \to 0} \frac{{h}^{\frac{1}{3}}}{h} = {\lim}_{h \to 0} \frac{1}{{h}^{\frac{2}{3}}}$ does not exist. This also implies that $f ' ' \left(0\right)$ does not exist.

Even though $f ' \left(0\right)$ does not exist, the graph of $f$ does have a vertical tangent line at $x = 0$ and you can see visually that there is an inflection point there as well in the picture below.

graph{x^(1/3) [-5, 5, -2.5, 2.5]}