At what value of x do the graphs of y=x^(2) and y=-sqrt (x) have perpendicular tangent lines?
Let m1 = slope of y = x^2
i.e. m1 = y' = 2x
Let m2 = slope of y = -sqrt(x)
i.e. m2 = y' = -(1/2)/sqrt(x)
These two slopes will be perpindicular when m1*m2 = -1
So, (2x)*(-(1/2)/sqrt(x)) = -1
-x/sqrt(x) = -1
x = sqrt(x)
The solution here is: x = 0 or x = 1