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

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Answer 1 · 2017-12-11T16:19:45

Please see below.

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

For $f(x) = x^(1/3)(x+3)^(2/3)$ , we have

Domain of $f$ is $(-oo,oo)$ and

$f'(x) = 1/3x^(-2/3)(x+3)^(2/3) + x^(1/3) 2/3(x+3)^(-1/3)$

$= (x+3)^(2/3)/(3x^(2/3)) + (2x^(1/3))/(3(x+3)^(1/3))$

Get a common denominator and combine to make one quotient.

$= ((x+3)+2x)/(3x^(2/3)(x+3)^(1/3))$

$= (x+1)/(x^(2/3)(x+3)^(1/3))$

$f'(x) = (x+1)/(x^(2/3)(x+3)^(1/3))$ is $0$ at $x=-1$

and $f'(x)$ 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]}

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