- #1

coolusername

- 36

- 0

## 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