Question

What are the local extrema, if any, of `f(x) =x^2(x+2)`?

Answer 1

`x=0, -4/3`

Explanation:

Find the derivative of `f(x)=x^2(x+2)`.
You will have to use the product rule.

`f'(x)=x^2+(x+2)2x=x^2+2x^2+4x=3x^2+4x`
`f'(x)=x(3x+4)`

Set `f'(x)` equal to zero to find the critical points.

`x=0`
`3x+4=0 rarr x=-4/3`

`f(x)` has local extrema at `x=0, -4/3`.
OR
`f(x)` has local extrema at the points (0, 0) and (`-4/3`, `32/27`).