1 answer
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a) x=0, x=2
b)
f(x) = x^3(x-2)^2
f '(x) = 3x^2(x-2)^2 + 2x^3(x-2)
= x^2(x-2)(5x-6)
critical points at x=0,2,6/5
max/min are at (0,0)(2,0)(1.2,1.106)
inflection where f ''(x) = 0
f ''(x) = 4x(5x^2 - 12x + 6)
f ''(x) = 0 at x = .71, 1.69
plug in to obtain f(x) there
concave down: -oo < x < 0
concave up: 0 < x < .71
concave down: .71 < x < 1.69
concave up: 1.69 < x < oo