# Question be380

Jan 15, 2017

0, at $x = 2 \mathmr{and} x = - 1$ and 16 at $x = - 2$, with explanation and graphical depiction.

#### Explanation:

f = 0, at x = 0, 0, 0, 2 and 2.

$f ' = {\left(x + 1\right)}^{2} \left(x - 2\right) \left(5 x - 4\right) = 0$, at x = 0, 0, 3 and 0.8.

$f ' ' = 2 \left(x + 1\right) \left(' 10 {x}^{2} - 12 x + 1\right) = 03$, at x$= 0 , 1.2 \pm 0.4 \sqrt{23}$

$f ' ' ' \ne 0$ at x = 0.

$| f |$ is the minimum 0 at $x = 2 \mathmr{and} x = - 1$ and

the maximum 16, at $x = - 2$.

Note that, at the turning point x = 0.8, f(.8) = 4.7 < 16.

See the graph, depicting all these aspects.

graph{(x+1)^3(x-2)^2 [-5, 5, -20, 20]}