# How do you find the exact relative maximum and minimum of the polynomial function of f(x)= x^2(x+2)?

Jan 4, 2018

Local maxima and minima are located at the point on the function where the first derivative is equal to zero.

#### Explanation:

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

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

Thus, the local extrema are at:
$f ' \left(x\right) = 0$ at $x = 0 \mathmr{and} x = - \frac{2}{3}$

Whether they are maxima or minima can be found from from the value of the first derivative on each side. The concavity or convexity can be determined from the sign of the second derivative at the inflection point,

$f ' ' \left(x\right) = 6 x + 2$
At $x = 0$
$f ' ' \left(x\right) = 2$ (positive) convex
and
at $x = - \frac{2}{3}$
$f ' ' \left(x\right) = - 2$ (negative) concave