# How do you find the x coordinates of all points of inflection, final all discontinuities, and find the open intervals of concavity for y=-(x+2)^(2/3)?

Jan 12, 2018

#### Explanation:

Let $f \left(x\right) = - {\left(x + 2\right)}^{\frac{2}{3}}$ and note that $f \left(x\right) = - \sqrt[3]{{\left(x + 2\right)}^{2}}$.

$f$ has domain: all real numbers and is continuous everywhere.

To investigate concavity look at $f ' ' \left(x\right)$

$f ' \left(x\right) = - \frac{2}{3} {\left(x + 2\right)}^{- \frac{1}{3}}$ and

$f ' ' \left(x\right) = \frac{2}{9} {\left(x + 2\right)}^{- \frac{4}{3}} = \frac{2}{9} \frac{1}{\sqrt[3]{{\left(x + 2\right)}^{4}}}$.

Since $f ' ' \left(x\right)$ is positive for $x \ne - 2$ (it is undefined at $- 2$), the graph of $f$ is concave up on $\left(- \infty , - 2\right)$ and also on $\left(- 2 , \infty\right)$

Since the concavity does not change, there is no point of inflection.

Here is the graph:
graph{-(x+2)^(2/3) [-10, 10, -5, 4.995]}