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

## 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

## The Attempt at a Solution

Related Calculus and Beyond Homework Help News on Phys.org
LCKurtz
Homework Helper
Gold Member
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?

• 1 person
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?

Ray Vickson
Homework Helper
Dearly Missed

## 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

## The Attempt at a Solution

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

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

Thanks LCKurtz & Ray Vickson!

LCKurtz