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

  • #1

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

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
LCKurtz
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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?
 
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  • #3
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
Ray Vickson
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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

Homework Statement





Homework Equations





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##.
 
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  • #5
Oh, I forgot to do the product rule on fy thus my fyy was incorrect.

Thanks LCKurtz & Ray Vickson!
 
  • #6
LCKurtz
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755
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.
 

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