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

Nov 7, 2017

$x = 0 , - \frac{4}{3}$

Explanation:

Find the derivative of $f \left(x\right) = {x}^{2} \left(x + 2\right)$.
You will have to use the product rule.

$f ' \left(x\right) = {x}^{2} + \left(x + 2\right) 2 x = {x}^{2} + 2 {x}^{2} + 4 x = 3 {x}^{2} + 4 x$
$f ' \left(x\right) = x \left(3 x + 4\right)$

Set $f ' \left(x\right)$ equal to zero to find the critical points.

$x = 0$
$3 x + 4 = 0 \rightarrow x = - \frac{4}{3}$

$f \left(x\right)$ has local extrema at $x = 0 , - \frac{4}{3}$.
OR
$f \left(x\right)$ has local extrema at the points (0, 0) and ($- \frac{4}{3}$, $\frac{32}{27}$).