Question

How do you find the local extremas for `f(x)=x^(1/3)(x+8)`?

Answer 1

Has a local minimum at `x = -2`

Explanation:

The critical points are obtained by solving
`d/(dx)f(x)=(4 (2 + x))/(3 x^(2/3))=0`
Solving for `x` we get `x = -2`
The critical point qualification is done by calculating `d^2/(dx)^2 f(x) = (4 (-4 + x))/(9 x^(5/3))`
So `d^2/(dx)^2 f(-2) = 0.839947`
Then the critical point is a minimum.
If you where using a symbolic processor be aware of
`x^{1/3} equiv x/(abs x)abs x^{1/3}` and `x^{5/3} equiv x/(abs(x))abs[x]^(5/3)`