# 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

## Answers and Replies

LCKurtz
Science Advisor
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
Science Advisor
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
Science Advisor
Homework Helper
Gold Member
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?

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.