1. find each value of x at which$\displaystyle f(x) =x(2x-8)^3$has a horizontal tangent line.

$\displaystyle y'=(2x-8)^3 + 6x(2x-8)^2$

set = to 0

$\displaystyle (2x-8)^3 + 6x(2x-8)^2=0$

factoring out

=$\displaystyle [(2x-8)^2][(2x-8)+(6x)]$

=$\displaystyle (2x-8)=0$

x=4

=$\displaystyle (6x)=0$

x=0

Key says the answer is x=1, and x=4.