Question

Question #be380

Answer 1

0, at `x = 2 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'=(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]}