Question

find the points of inflection for 2x(x-4)^3

Answer 1

Look at where f '' (x) = 0.

f(x) = 2x (x - 4)^3
f ' (x) = 2 (x - 4)^3 + 2x * 3(x - 4)^2
f ' (x) = 2(x - 4)^3 + 6x(x - 4)^2
f '' (x) = 2*3(x - 4)^2 + 6(x - 4)^2 + 6x*2(x - 4)
f '' (x) = 12(x - 4)^2 + 12x(x - 4)
f '' (x) = 12(x - 4)(x - 4 + x)
f '' (x) = 12(x - 4)(2x - 4)

12(x - 4)(2x - 4) = 0
(x - 4)(2x - 4) = 0
x - 4 = 0 or 2x - 4 = 0
x = 4 or 2x = 4
x = 4 or x = 2

Check x = 2:
f '' (1) = 12(1 - 4)(2*1 - 4) = 12(-3)(-2) > 0
f '' (3) = 12(3 - 4)(2*3 - 4) = 12(-1)(2) < 0
Since there is a change in sign around x = 2, this is an inflection point
Note that f(2) = 2*2 (2 - 4)^3 = 4(-2)^3 = 4 * -8 = -32

Check x = 4:
f '' (3) = 12(3 - 4)(2*3 - 4) = 12(-1)(2) < 0
f '' (5) = 12(5 - 4)(2*5 - 4) = 12(1)(6) > 0
Since there is a change in sign around x = 4, this is an inflection point
Note that f(4) = 2*4 (4 - 4)^3 = 8 * 0^3 = 8 * 0 = 0

Points of inflection: (2, -32) and (4, 0)

Answer 2

Refer to a Mathematica solution attached.
The results seem to be the same as @ilovenyc's .