Question

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

Answer 1

Please see below.

Explanation:

`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]}