
Kenneth S. answered 01/19/18
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Expert Help in Algebra/Trig/(Pre)calculus to Guarantee Success in 2018
y = x(x-2)3
y' = 1(x-2)3 + 3x(x-2)2 = (x-2)2[(x-2) + 3x] = 0
(x-2)2(2)(2x-1) = 0 has critical points when x=2 or x=½.
at critical number 2, the sign of y' is positive just to the left and also just to the right, so we have this behavior of y' over the numberline:
on (-infinity,½) y' is negative, y is decreasing
on (½,2) y' is positive, y is increasing
on (2,infinity) y' is positive, y is increasing
This information tells us that y has a local minimum at x=½.
I leave it to you to determine if there's an inflection point. Also, you may find the y coordinates as needed.