# How do you find the critical numbers for f(x) = x^(1/3)*(x+3)^(2/3) to determine the maximum and minimum?

Dec 11, 2017

#### Explanation:

$c$ is a critical number for $f$ if and only if $c$ is in the domain of $f$ and either $f ' \left(c\right) = 0$ or $f ' \left(c\right)$ does not exist.

For $f \left(x\right) = {x}^{\frac{1}{3}} {\left(x + 3\right)}^{\frac{2}{3}}$, we have

Domain of $f$ is $\left(- \infty , \infty\right)$ and

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

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

Get a common denominator and combine to make one quotient.

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

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

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

and $f ' \left(x\right)$ does not exist for $x = - 3$, and $x = 0$

These are all in the domain of $f$, so the critical numbers are:

$x = - 3$ (local max)
$x = - 1$ (local min)
$x = 0$ (neither min nor max)

The graph of $f$ is shown below.

graph{ x^(1/3)(x+3)^(2/3) [-7.024, 7.02, -3.51, 3.514]}