Question

How do you find the relative extrema for `f(x) =2x- 3x^(2/3) +2` on the interval [-1,3]?

Answer 1

Find and test the critical numbers for `f`.

`f(x) =2x- 3x^(2/3) +2`

Critical numbers for `f` are values for `x` that are:

1. In the domain of `f`, and

2. at which `f'(x) = 0` or `f'(x)` does not exist.

(This is a wonderful example of why you cannot ignore the second kind of critical numbers.)

For this function, we have:

`f'(x) = 2-2x^(-1/3) = 2(1-x^(-1/3)) = 2((root(3)x -1)/root(3)x)`

`f'(x) = 0` `color(white)"sssssssss"` `f'(x)` does not exist

`x=1` `color(white)"ssssssssssssss"` `x =0`

Using the first derivative we find that `f` is

increasing on `(-oo, 0)` (Yes, I know the domain has been restricted, but it is not necessary.)

decreasing on #(0,1) so

`f(0) = 2` is a relative maximum.

`f` is increasing on #(1,oo), so

`f(1) = 1` is a relative minimum.