Find the point where the curve y =e^-x has maximum curvature.

Answer 1

k = |y"|/(1+y'^2)^(3/2)
so,
k = e^-x/(1+e^2x)^(3/2)
find where k' = 0 for max curvature. Just thinking of the curve, where do you estimate it has max curvature? I'd guess somewhere near x = -3/2
Let's see what it really is.
k' = e^-x (e^2(2x+3) + 2) / 2(1+e^2x)^(5/2)
So, we want
e^2(2x+3) + 2 = 0
x = (-2/e^2 - 3)/2 = -(1/e^2 + 3/2) ≈ -1.635