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Homework Help: Finding Saddle point for a function

  1. May 30, 2012 #1
    I have the function:
    f(x,y)= x*(y^2)*e^-((x^2+y^2)/4)
    I found the critical points by first taking:
    partial with respect to x: e^(-(x^2+y^2)/4)*((y^2)-.5(x^2)(y^2))
    partial with respect to y:e^(-(x^2+y^2)/4)*(2yx-.5x(y^3))
    solving for 0 the critical points I got were:
    (√2,2),(√2,-2),(-√2,2),(-√2,-2) and (x,0) (for any value x there is a min or max along the y axis)
    The first 2 are local max, second 2 are local min, and the 5th point has both min and max.
    But I am confused, what is the saddle point for this function?
  2. jcsd
  3. May 31, 2012 #2
    Questions such as local extrema and saddle points are usually answered using the second derivative test. Does this sound right?

    (Global extrema are usually answered using simple comparison at critical points.)

    What do you mean by point has both max and min?
  4. May 31, 2012 #3
    If you look at the function along the y axis, when x is less than 0, the function is a local maximum and when x is more than 0 the function is a local minimum. This is looking only at the points on y=0. It looks a bit like a cylinder cut in half.
    I used the second derivative test and found the minimum and maximum values, but I can't find a saddle point.
  5. May 31, 2012 #4


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    Apparently, this function has no saddle points. The function has a constant value of zero along the y-axis .
  6. May 31, 2012 #5
    So the second derivative test gave you zero at your max min point?

    If this is the case, then the second derivative test would be inconclusive there (when zero). In these cases, we may be able to find other ways to classify the point. In fact, it sounds like you resourcefully accomplished this. Since it is a max along one slice, and a min along another slice, it is not a local extrema at all. Then I beleive it falls under the defintion of a saddle point, which I just read is a critical point which is not a local extrema.
  7. May 31, 2012 #6
    When I solved for the critical point, I got (x,0) as one of the points. looking at the graph I saw that is a min and max. But I am confused about how a whole line of points could be considered a saddle point? Or would (0,0) be the saddle point in this case, because if you put in (0,0) as a critical point and solve you get D=0, inconclusive.
  8. May 31, 2012 #7
    Logic of steps:

    Find critical points.

    Apply second derivative test

    • D>0, local extrema
    • D<0, saddle point
    • D=0, inconclusive

    Inconclusive does not imply saddle point, for example, try x^4+y^4, it has critical point at x,y=0, second derivitave is zero, but it is a minumum.

    The definition I saw of saddle point, is critical point that is not a local extrema.
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