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Answer 1 · 2017-01-15T14:44:01

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

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

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

$f''=2(x+1)('10x^2-12x+1)=03$ , at x $= 0, 1.2+-0.4sqrt23$

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

$|f|$ is the minimum 0 at $x = 2 and x = -1$ 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]}