# How do you determine where the given function f(x) = (x+3)^(2/3) - 6 is concave up and where it is concave down?

May 1, 2015

In order to investigate concavity, we'll look at the sign of the second derivative.

$f \left(x\right) = {\left(x + 3\right)}^{\frac{2}{3}} - 6$

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

Notice that $x = - 3$ there is a cusp at which the tangent becomes vertical. (The derivative goes to $\infty$)

f''(x) = -2/9(x+3)^(-4/3) = (-2)/(9(x+3)^(4/3)) = (-2)/(9(root(3)(x+3)^4)

The only place where $f ' '$ might change sign is at $x = - 3$.

But clearly the numerator is always negative, and the denominator, being a positive times a 4th power, is always positive.
So $f ' '$ is always negative where it is defined.

The graph is concave down on $\left(- \infty , - 3\right)$ and on $\left(- 3 , \infty\right)$.

Because of the cusp at x=0#, we cannot combine these two intervals.

Here's the graph of $f$

graph{(x+3)^(2/3) - 6 [-18.74, 13.3, -15.11, 0.92]}