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Basic maths bloody problem!

  1. Sep 11, 2004 #1
    i have been give the equation

    f(x) = 1/(x+3) + 1/(x-1)

    i am trying to find the stationary points. so

    f '(x) = -1/(x+3)^2 -1/(x-1)^2

    f '(x) = 0 for stationary point

    then when i simplified down i get 2x^2+4x+5 = 0, which is undefinable

    i looked at the derivative on the graph and it looks like it has a turning point at (-1,0).

    please help, either i have made an easy mistake or somethings wrong! :cry:
     
  2. jcsd
  3. Sep 11, 2004 #2
    Yeah, looks like the derivative has a "turning point" around x = -1. However, that doesn't mean that the derivative IS zero at that point, it means that the derivative of the derivative is zero there. If you wanted to x such that f'(x) = 0, you'd look for the places where f'(x) crosses the x-axis (which, in this case, it doesn't appear to do).
     
  4. Sep 11, 2004 #3
    Shouldn't it be 2x^2 + 4x + 10 =0. this has no real solution so the function is always increasing.

    You gotta look at the second derivative (f'') of this function and solve f'' = 0. Then you will get the x-coördinate of the turning point. Substituting this x-value into f(x) yields the y-value.


    regards
    marlon
     
  5. Sep 11, 2004 #4

    arildno

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    marlon:
    The function is always decreasing, not increasing!
    The function doesn't have any stationary points.
    The function has three continuous sections , each of them decreasing; the intervals where the function is defined are:
    [tex]-\infty<{x}<{-3},-3<{x}<{1},1<x<\infty[/tex]
     
  6. Sep 11, 2004 #5
    yes that is correct. Forgot the minus sign in the first derivative...

    Thanks for the correction.

    regards
    marlon
     
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