# How do you find points of inflection and determine the intervals of concavity given y=(x+2)^(1/3)?

Jan 8, 2018

See below.

#### Explanation:

Finding the first derivative and solving for $0$ will identify any stationary points. These can be max/min of points of inflection.

$\frac{d}{\mathrm{dx}} {\left(x + 2\right)}^{1} / 3 = \frac{1}{x + 2} ^ \left(\frac{2}{3}\right)$

$f ' \left(x\right) = 0 \implies \frac{1}{x + 2} ^ \left(\frac{2}{3}\right) = 0$

$0 = 1$

This has no solutions. ( no max/min, but possible inflection point )

Using second derivative.

$f ' ' \left(x\right) > 0$ convex ( concave up )

$f ' ' \left(x\right) < 0$ concave (concave down )

Second derivative:

$f ' ' \left(x\right) = f ' \left(f ' x\right)$

$\frac{d}{\mathrm{dx}} \left(\frac{1}{x + 2} ^ \left(\frac{2}{3}\right)\right) - \frac{2}{9 {\left(x + 2\right)}^{\frac{5}{3}}}$

$- \frac{2}{9 {\left(x + 2\right)}^{\frac{5}{3}}} < 0$

$9 {\left(x + 2\right)}^{\frac{5}{3}} > 0$

$x + 2 > 0$

$x > - 2$

Concave down $\textcolor{b l u e}{\left[- 2 , \infty\right)}$

$- \frac{2}{9 {\left(x + 2\right)}^{\frac{5}{3}}} > 0$

$\frac{1}{{\left(x + 2\right)}^{\frac{5}{3}}} < 0$

$\left(x + 2\right) < 0$

$x < - 2$

Convex $\textcolor{b l u e}{\left(- \infty , - 2\right]}$

Inflection point at $\textcolor{b l u e}{\left(\left(- 2 , 0\right)\right)}$

GRAPH:

graph{y=(x+2)^(1/3) [-12.66, 12.65, -6.33, 6.33]}