Question

What are the critical points of `f(x)=x^(2/3)x^2 - 16`?

Answer 1

`c=0`

Explanation:

Critical points (`c`) occur when `f'(c)=0` or when `f'(c)` doesn't exist.

In order to find `f'(x)`, first simplify `f(x)`.

`f(x)=x^(2/3)x^2-16`

`f(x)=x^(2/3)x^(6/3)-16`

`f(x)=x^(8/3)-16`

Use the power rule: `d/dx[x^n]=nx^(n-1)`

`f'(x)=8/3x^(5/3)`

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

Thus, `c=0`.

graph{x^(2/3)x^2-16 [-32.06, 32.9, -20.27, 12.2]}