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Maximums and Minimums

  1. Jun 18, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the absolute max/min of the function on it's interval.
    [tex]f(x)=\frac{x^2-4}{x^2+4}[/tex]
    Interval: [-4,4]

    2. Relevant equations

    3. The attempt at a solution
    [tex]f(x)=\frac{x^2-4}{x^2+4}[/tex]

    I basically want to find all the critical points, so I set the denominator to zero and found a critical point to be where x = 2i, and x = -2i.

    Then I took the derivative of the function as so:

    [tex]f'(x) = \frac{2x(x^2+4)-2x(x^2-4)}{(x^2+4)^2}[/tex]

    Setting the numerator to zero should find where the derivative is equal to zero, but that expands out to this:
    [tex]2x^3+8x-2x^3+8x[/tex]
    Which is basically zero anyway. I was going to try to factor out something from the numerator and denominator (if possible) to cancel it out so I could get something, but I thought I would lose solutions doing that. So here is where I am confused? I tried graphing it, thinking that it would just return the line y=0, but my TI-89 just says "not a function or program, error"?

    edit: fixed
     
    Last edited: Jun 18, 2011
  2. jcsd
  3. Jun 18, 2011 #2
    Hint Hint: 2 negatives make a positive. Look at your expansion again :)
     
  4. Jun 18, 2011 #3
    Ah yeah. Thanks! So the other critical point is 0 then.
     
  5. Jun 18, 2011 #4

    Dick

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    Your algebra is faulty. The x terms don't cancel. And you probably don't care about imaginary solution in the denominator.
     
  6. Jun 18, 2011 #5
    [tex]2x^3+8x-2x^3+8x=0[/tex]
    from the numerator, works out to be:
    [tex]16x=0[/tex]
    x=0

    I was not sure what to do with the imaginary solutions, essentially the only critical point found is at x=0 then I would imagine. Leaving me to check the endpoints and x=0 so find the absolutes correct?
     
  7. Jun 18, 2011 #6

    Dick

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    Correct. The imaginary solutions are irrelevant. They aren't in the real interval [-4,4].
     
  8. Jun 18, 2011 #7
    Alright, I think I got it now. This was the only one in the set of problems that had the possibility of an imaginary solution, and I really wasn't sure if that counted or not.

    Thanks for the help.
     
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