find the x coordinate of each point at which the line tangent to the graph of f(x)=x^4-3x^2 is parallel to the line y= -2x+4

Answer 1

f ' (x) = 4x^3 - 6x

since parallel to y = -2x+4 , slope must be -2

4x^3 - 6x = -2
2x^3 - 3x + 1 = 0
try the factor theorem ...
x = 1, LS = 2-3+1 = 0 = RS
so x-1 is a factor
2x^3 - 3x + 1 = (x-1) (2x^2 + 2x - 1) = 0
x = 1
or
2x^2 + 2x-1 = 0
x = (-2 ± √12)/4
= (-1 ± √3)

x = 1, (-1+√3)/2 , (-1-√3)/2