- #1
coolusername
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Homework Statement
The function f(x,y) = [e^(-y^2)]cos(4x) has a critical point (0,0)
Homework Equations
Find the D value at the critical point. What type of critical point is it? (max, min, saddle or none)
The Attempt at a Solution
I know that to find the D value I must compute the partial derivatives (fox, fey, fxy and fyx)
fx = [-4e^(-y^2)]sin(4x)
fy = [-2ye^(-y^2)]cos(4x)
fxx = [-16e^(-y^2)](cos(4x)
fyy = [4(y^2)e^(-y^2)]cos(4x)
fxy = [8ye^(-y^2)]sin(4x)
fxy = [8ye^(-y^2)]sin(4x)
D = (fxx)(fyy) - (fxy)^2
= [-16e^(-y^2)](cos(4x)[4(y^2)e^(-y^2)]cos(4x) - {[8ye^(-y^2)]sin(4x)}^2
evaluated at the critical point (0,0)
= (-16)(0) - (0)(0)
= 0
This means that the 2nd derivative test gave no info. But somehow it's not the right answer. Did I find the correct value of D? Is my approach correct?
Thanks