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Math Help - Stationary points with 2 variables

  1. #1
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    Stationary points with 2 variables

    Hi,

    Locate and classify all of the stationary points of the function:

    f(x,y) = e^x(3x - y^3 + 3y).

    I have got as far as doing the partial derivatives for the terms but I am unsure if the partial derivatives I got are right

    What I got for the derivatives so far are:

    f_{x} = 3e^x - e^xy^3
    f_{xx} = 3e^x - e^xy^3 <-- unsure on this one
    f_{y} = 3xe^x - 3e^xy^2 + 3
    f_{yy} = 3xe^x - 6e^xy
    f_{xy} = 3e^x - e^xy^3

    I am sure that after I have found these I am meant to equate them to zero (because they are stationary points) and then find x and y and then put the in one of the equations (I think its the original one if I remember correctly) and get the coordinates of the stationary points.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Beard View Post
    Hi,

    Locate and classify all of the stationary points of the function:

    f(x,y) = e^x(3x - y^3 + 3y).

    I have got as far as doing the partial derivatives for the terms but I am unsure if the partial derivatives I got are right

    What I got for the derivatives so far are:

    f_{x} = 3e^x - e^xy^3
    f_{xx} = 3e^x - e^xy^3 <-- unsure on this one
    f_{y} = 3xe^x - 3e^xy^2 + 3
    f_{yy} = 3xe^x - 6e^xy
    f_{xy} = 3e^x - e^xy^3

    I am sure that after I have found these I am meant to equate them to zero (because they are stationary points) and then find x and y and then put the in one of the equations (I think its the original one if I remember correctly) and get the coordinates of the stationary points.
    f_x=\frac{\partial}{\partial x}\left[ 3x e^x + (-y^3+3y)e^x\right]=3e^x+3xe^x+(-y^3+3y)e^x

    CB
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  3. #3
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    I found where I was going wrong for the differentiations, I wasn't using the product rule when I did it.

    I have redone the differentials using the rule and come up with this.

    f_x = 3e^x + e^x(3x - y^2 +3y)
    f_{xx} = 6e^x + e^x(3x - y^2 +3y)
    f_{xy} = 6 + 3x + y + y^2
    f_y = e^x(-2y +3) + 3x - y^2 + 3y
    f_{yy} = -2e^x - 2y +3

    I tried equating one of the to zero but I'm not getting anywhere and keep hitting dead ends. Am I able to divide by e^x on some of them?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Beard View Post
    I found where I was going wrong for the differentiations, I wasn't using the product rule when I did it.

    I have redone the differentials using the rule and come up with this.

    f_x = 3e^x + e^x(3x - y^2 +3y)
    f_{xx} = 6e^x + e^x(3x - y^2 +3y)
    f_{xy} = 6 + 3x + y + y^2
    f_y = e^x(-2y +3) + 3x - y^2 + 3y
    f_{yy} = -2e^x - 2y +3

    I tried equating one of the to zero but I'm not getting anywhere and keep hitting dead ends. Am I able to divide by e^x on some of them?
    f_y is still wrong

    CB
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