Question

What are the global and local extrema of `f(x) = x^2(2 - x)` ?

Answer 1

`(0,0)` is a local minimum and `(4/3,32/27)` is a local maximum.
There are no global extrema.

Explanation:

First multiply the brackets out to make differentiating easier and get the function in the form
`y=f(x)=2x^2-x^3`.

Now local or relative extrema or turning points occur when the derivative `f'(x)=0`,
that is, when `4x-3x^2=0`,
`=> x(4-3x)=0`
`=>x=0 or x=4/3`.

`therefore f(0)=0(2-0)=0 and f(4/3)=16/9(2-4/3)=32/27`.

Since the second derivative `f''(x)=4-6x` has the values of
`f''(0)=4>0 and f''(4/3)=-4<0`,
it implies that `(0,0)` is a local minimum and `(4/3,32/27)` is a local maximum.

The global or absolute minimum is `-oo` and the global maximum is `oo`, since the function is unbounded.

The graph of the function verifies all these calculations :

graph{x^2(2-x) [-7.9, 7.9, -3.95, 3.95]}