Question #be380

1 Answer
Jan 15, 2017

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]}