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

Mar 25, 2016

$\left(0 , 0\right)$ is a local minimum and $\left(\frac{4}{3} , \frac{32}{27}\right)$ 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
$y = f \left(x\right) = 2 {x}^{2} - {x}^{3}$.

Now local or relative extrema or turning points occur when the derivative $f ' \left(x\right) = 0$,
that is, when $4 x - 3 {x}^{2} = 0$,
$\implies x \left(4 - 3 x\right) = 0$
$\implies x = 0 \mathmr{and} x = \frac{4}{3}$.

$\therefore f \left(0\right) = 0 \left(2 - 0\right) = 0 \mathmr{and} f \left(\frac{4}{3}\right) = \frac{16}{9} \left(2 - \frac{4}{3}\right) = \frac{32}{27}$.

Since the second derivative $f ' ' \left(x\right) = 4 - 6 x$ has the values of
$f ' ' \left(0\right) = 4 > 0 \mathmr{and} f ' ' \left(\frac{4}{3}\right) = - 4 < 0$,
it implies that $\left(0 , 0\right)$ is a local minimum and $\left(\frac{4}{3} , \frac{32}{27}\right)$ is a local maximum.

The global or absolute minimum is $- \infty$ and the global maximum is $\infty$, 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]}