# Find the D value of critical point and the type of critical point

1. Nov 28, 2013

1. The problem statement, all variables and given/known data

The function f(x,y) = [e^(-y^2)]cos(4x) has a critical point (0,0)

2. Relevant equations

Find the D value at the critical point. What type of critical point is it? (max, min, saddle or none)

3. 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
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 28, 2013

### LCKurtz

Having the discriminant be zero doesn't mean there is no max or min. In your case, $f(0,0)=1$. Can you tell from the original equation whether that might be a max or min?

3. Nov 28, 2013

I see from the original equation that the max of f(x,y) is 1. However what would the value of D be? Did I calculate it correctly? I keep getting 0 but since the critical point has a max behaviour then shouldn't D be more than 0?

4. Nov 28, 2013

### Ray Vickson

Your second derivative $f_{yy}$ is incorrect; it should not be 0 at (0,0), and $D(0,0) \neq 0$.

5. Nov 28, 2013

Oh, I forgot to do the product rule on fy thus my fyy was incorrect.

Thanks LCKurtz & Ray Vickson!

6. Nov 28, 2013

### LCKurtz

Regardless of the calculation error that I overlooked, the discriminant can be zero even though there is a relative max or min. This is even true in one variable. Consider $y=x^4$. This has an absolute min at $x=0$ yet $y'=0$ and $y''=0$ at $x=0$ so the second derivative test fails to discriminate the relative min.