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

  1. Nov 28, 2013 #1
    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. jcsd
  3. Nov 28, 2013 #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?
     
  4. Nov 28, 2013 #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?
     
  5. Nov 28, 2013 #4

    Ray Vickson

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    Your second derivative ##f_{yy}## is incorrect; it should not be 0 at (0,0), and ##D(0,0) \neq 0##.
     
  6. Nov 28, 2013 #5
    Oh, I forgot to do the product rule on fy thus my fyy was incorrect.

    Thanks LCKurtz & Ray Vickson!
     
  7. Nov 28, 2013 #6

    LCKurtz

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    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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