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]}