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Finding the critical point and its nature. With solid attempt

  1. Nov 2, 2011 #1
    Finding the critical point and its nature. With solid attempt!!

    1. The problem statement, all variables and given/known data

    Find all critical points of the function

    f(x, y) = xy2 - 2xy - 2x2 - 3x +7

    and determine their nature.

    2. Relevant equations

    none

    3. The attempt at a solution

    I know that to find the critical points you must set fx = 0 and fy=0

    Doing this I get:


    fx = -4x + y2 - 2y - 3 = 0

    and

    fy = 2xy - 2x = 0

    Thus, fx = fy

    -4x + y2 -2y - 3 = 0

    -4x + (y+1)(y-3) = 0

    I don't really know how to proceed? Any help would be great.
     
  2. jcsd
  3. Nov 2, 2011 #2

    Dick

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    Re: Finding the critical point and its nature. With solid attempt!!

    Factor the fy equation first. What conclusions can you draw from that?
     
  4. Nov 2, 2011 #3
    Re: Finding the critical point and its nature. With solid attempt!!

    fy = 2xy - 2x = 2x(y-1)

    Thus, x=0 and y=1

    SO, critical point is (0,1)...

    So is that the only critical point or do I sub this back into fx to get another?

    Thanks Dick!
     
  5. Nov 2, 2011 #4

    Dick

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    Re: Finding the critical point and its nature. With solid attempt!!

    Careful!! 2x(y-1)=0 if x=0 OR y=1. Not necessarily both. Put those two possibilities back into fx and see what happens.
     
  6. Nov 2, 2011 #5
    Re: Finding the critical point and its nature. With solid attempt!!

    Silly me!

    Okay subbing back into fx gives me: (0,-1), (0,3), (-1,1)

    Assuming these are correct do I now find fxx, fyy, and fxy
    to determine the nature (using theorem) ??

    On a side note how do you find fxy? I'm confused about that particular instance.

    Thanks Dick! I really appreciate your help
     
  7. Nov 2, 2011 #6

    Dick

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    Re: Finding the critical point and its nature. With solid attempt!!

    Sure, now use the second derivative test on those three critical points. fxy isn't hard to to find, just take your fx and differentiate with respect to y, OR take fy and differentiate with respect to x. You'll get the same thing.
     
  8. Nov 2, 2011 #7
    Re: Finding the critical point and its nature. With solid attempt!!

    Got it thanks!

    Now I have anther question:

    Find all critical points of the function: f(x, y) = y sin x + cos x and determine their nature.

    So we have fx = ycosx - sinx and fy = sinx

    hence, for

    fx we have y = sinx/cosx = tanx

    and

    fy we have sinx = 0

    Therefore we have x=0... when we sub this into y=tanx we get y=0

    Hence critical point is (0,0)

    Would this be reasonable?

    Regards
     
  9. Nov 2, 2011 #8

    Dick

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    Re: Finding the critical point and its nature. With solid attempt!!

    x=0 isn't the only solution to sin(x)=0. What are the others?
     
  10. Nov 2, 2011 #9
    Re: Finding the critical point and its nature. With solid attempt!!

    ∏, 2∏

    Now what :/

    edit: x=∏(n) for all n ε Z
     
  11. Nov 2, 2011 #10

    Dick

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    Re: Finding the critical point and its nature. With solid attempt!!

    Put x=pi*n into fx. What does that tell you about y?
     
  12. Nov 2, 2011 #11
    Re: Finding the critical point and its nature. With solid attempt!!

    As you put it into fx, sinx = 0... therefore you have fx = -y for odd n and fx = y for even n..

    Hence y=0 again for x=n*∏

    ?
     
  13. Nov 2, 2011 #12

    Dick

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    Re: Finding the critical point and its nature. With solid attempt!!

    Fine. So your critical points are (-pi,0), (0,0), (pi,0), (2pi,0)... right? (n*pi,0) where n is any integer.
     
    Last edited: Nov 2, 2011
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