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Partial derivative stationary point

  1. Feb 23, 2017 #1
    1. The problem statement, all variables and given/known data
    Hi guys, I am having real trouble with the function 10ii) I can take the derivatives, but I feel like I am missing something, with what I have done. I set $f_x=0$and $f_y=0$ but really cant seem to find away to solve, i keep getting (0,0) which when I plug into wolfram it come out with $$x=0,y=-1/2$$, $$x=-1,y=0$$ and $$x=1, y=-2$$
    So this make me think have I done the partial right?

    upload_2017-2-23_19-49-18.png

    2. Relevant equations


    3. The attempt at a solution

    $$\frac{\partial}{\partial x}=-2(x^2y-x-1)(2xy-1)-2(x^2-1)(2x)$$
    $$\frac{\partial}{\partial y}=-2(x^2y-x-1)(x^2)$$


    $$-2(x^2y-x-1)(2xy-1)-2(x^2-1)(2x)=0$$
    $$-2(x^2y-x-1)(x^2)=0$$

    but where do I go from here?

    Could someone please give advice, thanks in advance.



     
  2. jcsd
  3. Feb 23, 2017 #2

    BvU

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    Two hashes (#) to get inline ##\TeX##, two dollars ($) to get displayed math :smile:

    Check ##f_x## and ##f_y## (*); once OK, ##f_y## gives you something to substitute in ##f_x=0##

    (*) [edit] right, thanks Dick :smile: (I copied the 2 inside the [ ] o:) )
     
    Last edited: Feb 23, 2017
  4. Feb 23, 2017 #3

    Dick

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    The partials look ok. Think about the y derivative first. It's zero if i) x=0 OR y=?. Fill in the ? with an expression in x. Substitute that into the x derivative.
     
  5. Feb 24, 2017 #4

    epenguin

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    Not every function of x and y has a stationary point - it is always possible that your function has none.
    In that case you will not be able to find it! But then you have to prove it doesn't exist.
    In this case I think there is a stationary point.
    fy is nicely factorised, which is help. One of the factors is closely related to a bracket in fx which is another help.

    Not every condition for one of your derivatives to be 0 necessarily allows the the other to be 0. :oldwink:
    (Though even if it doesn't, your zeros of one of them still gives information about what the surface f looks like)
     
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