Question

How do you find the intervals of increasing and decreasing using the first derivative given `y=x^(2/3)-4`?

Answer 1

`y` is decreasing on `(-oo,0)` and increasing on `(0,oo)`. The minimum value of `y` is `-4` (at `x=0`).

Explanation:

For `y=f(x) = x^(2/3)-4`, the domain is all real numbers a,d

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

`f'(x)` is never `0` and is undefined when `x=0`.

We check the sign of `f'(x)` on both sides of `0`.

On `(-oo,0)` we have `f'(x) < 0`, so `f` is decreasing.

On `(0,oo)` we have `f'(x) > 0`, so `f` is increasing.