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Local Max & Min

  1. Mar 25, 2008 #1
    1. Find the local max and min of x/(1+x^2)

    3. The attempt at a solution

    Take the derivative of x/(1+x^2) and set it equal to zero.

    f'(x)= (1-x^2)/(1+x^2)^2 = 0

    This is where I am stuck. My first guess was to factor the top, resulting in (1-x)(1+x)/(1+x^2)^2 = 0, and then cancel the 1+x. But after that I'm still left with (1-x)/(1+x)^2 = 0. I know if you let x=0 you will get f'(x) = 0 because 0/anything = 0.

    Really, I'm not sure what I can do after I take the derivative.
     
  2. jcsd
  3. Mar 25, 2008 #2
    Good. Now stop.

    [tex]\frac {(1-x)(1+x)}{(1+x^2)^2} \neq \frac {1-x}{(1+x)^2}[/tex]
     
  4. Mar 25, 2008 #3
    Would I factor the bottom into (1+x^2)(1+x^2)?
     
  5. Mar 25, 2008 #4
    What's the use? That bugger is never going to equal 0 no matter what number you put in the denominator. Look at the numerator.
     
  6. Mar 25, 2008 #5
    [tex]f'(x)= \frac{(1-x^2)}{(1+x^2)^2} = 0[/tex]

    Is 0 when the numerator is 0...

    [tex]1-x^2 = 0[/tex]

    [tex]x = \pm 1?[/tex]
     
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