Find the x-coordinates of all points on the curve y = (-x^2 + 4x - 3)^3 with a horizontal tangent line. (Thanks!)
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Originally Posted by Affinity Find the x-coordinates of all points on the curve y = (-x^2 + 4x - 3)^3 with a horizontal tangent line. (Thanks!) That is where, $\displaystyle y'=0$ Thus, using the chain rule, $\displaystyle 3(-2x+4)(-x^2+4x-3)^2=0$ Thus, $\displaystyle -2x+4=0$ thus, $\displaystyle x=2$ $\displaystyle -x^2+4x-3=0$ thus, $\displaystyle x=1,3$
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