Determine where (x-value only) on the curve y=1/(2x−1)^2 where the tangent line will be perpendicular to the line defined by 2x+y−1=0.
we haven't learned this in the curriculum so damon i would love your help rn
well, 2x+y-1=0 has slope -1/2, so we want our tangent line to have slope 2
y = 1/(2x-1)^2
y' = -4/(2x-1)^3
so, where is y'=2?
-4/(2x-1)^3 = 2
(2x-1)^3 = -2
x = (1-∛2)/2 ≈ -0.13