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Second derivative test for functions of 2 variables

  1. Oct 2, 2011 #1
    urgent! second derivative test for functions of 2 variables

    1. The problem statement, all variables and given/known data
    f(x,y)=x^4 - y^2 - 2x^2 + 2y - 7


    2. Relevant equations
    classify points (0,1) and (-1,1) as local maximum, local minimum or inclusive


    3. The attempt at a solution

    f(x,0)=4x^3 - 0 - 4x + 0 - 0 = 4x^3-4x
    f'(x,0)=12x^2-4
    f''(x,0)=24x

    f(0,y)=0 - 2y - 0 + 2 - 0 = -2y+2
    f'(0,y)=-2
    f''(0,y)=0

    how to find the points as above? as i am stuck.
     
  2. jcsd
  3. Oct 2, 2011 #2
    Re: urgent! second derivative test for functions of 2 variables

    Why are you fixing x=0 and y=0 when you calculate the derivates? You need to use the Hessian matrix to solve your problem.
     
  4. Oct 2, 2011 #3
    Re: urgent! second derivative test for functions of 2 variables

    so i use the 2x2 matrix?
    do i equate the 4x^3-4x=0 & -2y+2=0?

    sry as i am new just start learning.
     
  5. Oct 2, 2011 #4
    Re: urgent! second derivative test for functions of 2 variables

    No worries. Slow down though. Why are you equating anything to zero? Can you tell me what the definition of the Hessian is? If you can, calculate the components and put them into the matrix. After that, tell me how we can use the Hessian to determine whether a point is maximal, minimal, or a saddle?
     
  6. Oct 2, 2011 #5
    Re: urgent! second derivative test for functions of 2 variables

    i thought of finding the stationary point.
    square matrix of second-order partial derivatives.
    fx=4x^3-4x
    fy=-2y+2
    is this correct?
     
  7. Oct 2, 2011 #6
    Re: urgent! second derivative test for functions of 2 variables

    Yep. But now you need fxx fxy and fyy.
     
  8. Oct 2, 2011 #7
    Re: urgent! second derivative test for functions of 2 variables

    You can find the stationary points if you want, but I believe they've already been given to you.
     
  9. Oct 2, 2011 #8
    Re: urgent! second derivative test for functions of 2 variables

    fxx=12x^2-4
    fyy=-2
    fyx=0
    fxy=0

    correct?
     
    Last edited: Oct 2, 2011
  10. Oct 2, 2011 #9
    Re: urgent! second derivative test for functions of 2 variables

    Throw a negative sign in front of fyy and that will be correct. Now, how can you determine whether or not (0,1) or (-1,1) are min/max/saddle using the Hessian?
     
  11. Oct 2, 2011 #10
    Re: urgent! second derivative test for functions of 2 variables

    [12x^2-4 0]
    -2 0

    do i need to find the eigenvalues?
     
  12. Oct 2, 2011 #11
    Re: urgent! second derivative test for functions of 2 variables

    0,1 inconclusive -1,1 maximum?
     
  13. Oct 2, 2011 #12
    Re: urgent! second derivative test for functions of 2 variables

    anyone?
     
  14. Oct 2, 2011 #13
    Re: urgent! second derivative test for functions of 2 variables

    up!!
     
  15. Oct 2, 2011 #14
    Re: urgent! second derivative test for functions of 2 variables

    Check your matrix again, it should be

    [tex] \begin{pmatrix} 12x^2 & 0 \\ 0 & -2 \end{pmatrix} [/tex]

    Now you do indeed have to find the eigenvalues of this matrix for each point you want to consider. However, the eigenvalues of diagonal matrices are quite easy...
     
  16. Oct 3, 2011 #15
    Re: urgent! second derivative test for functions of 2 variables

    so 0,1 maximum -1,1 inconclusive
     
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