Question

How do you find the relative extrema for `f(x) =2x- 3x^(2/3) +2`?

Answer 1

Relative maximum `f(0) = 2` and relative minimum `f(1) = 1`

Explanation:

The domain of `f` is `RR`

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

Critical numbers for `f` are values of `x` in the domain of `f` at which `f'(x) = 0` or `f'(x)` does not exist.

`f'(x) = 0` at `x=1` and `f'(x)` fails to exist at `x=0`.
Both are in the domain of `f`, so both are critical numbers for `f`.

on the interval `(-oo,0)`, `f'(x)` is positive and on `(0,1)` it is negative, so `f(0)=2` is a relative maximum.

on `(1,oo)`, `f'(x)` is again positive, so `f(1) = 1` is a relative minimum.

Here is the graph of `f`:
graph{2x-3x^(2/3)+2 [-5.365, 5.735, -2.02, 3.53]}