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Maxima-minima problems

  1. Dec 17, 2004 #1
    Hi!

    For example y=-x^3-3x=0 gives y'=-3x^2-3 and setting y'=0 we get i and -i as the solutions. What does this say about the existence of the max and min points for the function y?

    - Kamataat
     
  2. jcsd
  3. Dec 17, 2004 #2
    Hi!

    -3x2-3=0 has no solutions in real numbers

    So, y' is always negative (as -3 is)
    Hence, y is always decreasing (no min and max)
     
  4. Dec 18, 2004 #3
    ok, thanks

    - Kamataat
     
  5. Dec 18, 2004 #4
    -x^3-3x=k

    -x^3-3x-k=0=> where b makes x only have two solutions.....
    x^3+3x-3x^2+k=(x^2-bx-(b/2)^2)(x+c)
    x^3-bx^2-b^4/4x+x^2c-bcx-cb^4/4
    k= -cb^4/4
    -b+c=0
    -b^4/4-b^2=-3
    b^4+4b^2=12
    b^4+4b^2-12=0
    (b^2-h)(b^2-a)
    (a+h)=-4
    ah=12
     
    Last edited: Dec 19, 2004
  6. Sep 17, 2010 #5
    hi!
    how determine whether that point is the maximum or the minimum?
     
  7. Sep 17, 2010 #6

    Mark44

    Staff: Mentor

    The second derivative test is helpful. At a critical number c for which f'(c) = 0, if f''(c) > 0, (c, f(c)) is a local minimum point; if f''(c) < 0, (c, f(c)) is a local maximum point.

    There's more to this, but your calculus text should have more information about the details.

    In the future, if you have a question, start a new thread rather than adding onto an old thread. This thread is six years old.
     
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