Find the points on y=(-x^2+4x-3)^3 that has a horizontal tangent line.

Answer 1

y' = 3(-x^2+4x-3)^2 (-2x+4)
y'=0 when x = 1,2,3
so, the tangent is horizontal at

(1,0),(2,1),(3,0)

Answer 2

Tnx! :D