# How do you determine the intervals where f(x)=x^(2/3)+3 is concave up or down?

Oct 25, 2016

Investigate the sign of the second derivative.

#### Explanation:

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

The domain of $f$ is all real numbers.

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

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

$f ' '$ is never $0$ and is undefined at $x = 0$.

So the only place where $f ' '$ might change signs is $x = 0$

On $\left(- \infty , 0\right)$, we see that $f ' '$ is negative, so $f$ is concave down.

On $\left(0 , \infty\right)$, we see again that $f ' '$ is negative, so $f$ is concave down.

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

By the way, here is the graph of $f \left(x\right)$.

graph{y=x^(2/3)+3 [-12.62, 12.69, -2.98, 9.68]}