CALC: How do I find the critical points of this function

f(x) = 2x(8-x)^3

I know I have to use the product rule to differentiate and then equal it to 0 and solve for x.
Somehow I'm having trouble trying to separate the x's so I can equal them to 0. Can somebody show me how to with steps?

thanks !

f(x) = 2x(8-x)^3
f '(x) = 2(8-x)^3 - 6x(8-x)^2 = (8-x)^2(16-2x - 6x) = - 8(8-x)^2(x-2)
critical points are 2 and 8

in order to find the critical points of the function, you have to take the derivative of f(x) and find where it equals zero. in order to do that, you have to use a combination of the product rule and the chain rule. in f(x), there are two terms: 2x and (8-x)^3. their derivatives are 2 and 3(8-x)^2 * -1 respectively. The derivative of (8-x)^3 can be obtained by the chain rule. Now when you use the product rule, you get

f '(x) = [2(8 - x)^3] + [-3(8 - x)^2 * 2x] = 2 (8 - x)^2 * [ (8 - x) +-3x] = 2 (8 - x)^2 [8 - 4x]

when you solve for f '(x) = 0, you get x = 8 and x = 2 as the critical points. to find the y values, plug them in for x into f(x) . i simplified f ' (x) by factoring out common terms

f'(x) = 2(8-x)³ + 6x(8-x)²(-1) = 0

2(8-x)³ - 6x(8-x)² = 0
(8-x)²[2(8-x) - 6x] = 0
(8-x)²(16-8x) = 0
(8-x)²(2-x) = 0

The critical points are at x = 8 and x = 2

(x = 2 is a relative maximum point, and x = 8 is an inflection point)

First differenciate f(x), then equals the result to zero and solve for x.