1. ## Short Calculus Problem!

Find the x-coordinates of all points on the curve y = (-x^2 + 4x - 3)^3 with a horizontal tangent line. (Thanks!)

2. 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, $\displaystyle x=1,3$