# How do you find points of inflection and determine the intervals of concavity given y=x^(1/3)(x-4)?

Jan 8, 2018

There is a point of inflection at $x = 2$. The graph is concave down for $x < 2$, and concave up for $x > 2$

#### Explanation:

Let's say $y = f \left(x\right) = {x}^{\frac{1}{3}} \left(x - 4\right) = {x}^{\frac{4}{3}} - 4 {x}^{\frac{1}{3}}$

This means:

$f ' \left(x\right) = \frac{4}{3} {x}^{\frac{1}{3}} - \frac{4}{3} {x}^{- \frac{2}{3}}$

and

$f ' ' \left(x\right) = \frac{4}{9} {x}^{- \frac{2}{3}} + \frac{8}{9} {x}^{- \frac{5}{3}} = \frac{4}{9} {x}^{- \frac{5}{3}} \left(x - 2\right)$

Points of inflection occur where f''(x) crosses the x-axis, which happens where f''(x) = 0, and there isn't a double root.

In the case of this problem, this occurs at $x = 2$.

By substituting numbers around 2, we can easily see that

$f ' ' \left(x\right) > 0 , x > 2$ and $f ' ' \left(x\right) < 0 , x < 2$

There is a point of inflection at $x = 2$. The graph is concave down for $x < 2$, and concave up for $x > 2$