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Absolute max and min

  1. Nov 6, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the absolute maximum and absolute minimum values of f on the given interval.
    f(x) = x/(x^2 + 25)
    [0, 8]


    2. Relevant equations
    taking derivatives are necessary


    3. The attempt at a solution

    [tex] f(x)= \frac{x}{x^2+25} [/tex]
    [tex] f'(x)= \frac{(x^2+25)-x(2x)}{(x^2+25)^2} [/tex]
    [tex] f'(x)= \frac{x^2+25-2x^2}{(x^2+25)^2} [/tex]
    [tex] f'(x)= \frac{-x^2+25}{(x^2+25)^2} [/tex]

    [tex] f'(x)= -x^2+25=0 [/tex]
    [tex] 25=x^2 [/tex]
    [tex] x=-5,5 [/tex]

    [tex] f(0)=0 [/tex] min
    [tex] f(5)=1/2 [/tex] max
    [tex] f(8)=8/89[/tex]

    Those are my answers but i think i am wrong...
     
  2. jcsd
  3. Nov 6, 2009 #2
    You can't cancel
    [tex] {-x^2+25} [/tex]
    and [tex]{(x^2+25)^2}[/tex]
    b/c of the negative.
    If you factor out the negative:
    [tex] \frac{-(x^2-25)}{(x^2+25)^2} [/tex]

    Which gives you a numerator of
    [tex] {-(x+5)(x-5)}[/tex]
    which are your zeros of f'(x) and you just plug those x-values of x = +/- 5 into f(x) and find your max and min.
     
  4. Nov 6, 2009 #3
    btw, you're x-values are right, but your process is flawed.
    Also, b/c of the interval, only f(5) applies from the derivative.

    You will also want to graph the equation to make sure.
    Graph all of f(x) and then eyeball where 0 and 8 are and just mark through them to show exactly the part of the curve with which you are dealing.
    You should do this with all functions, especially rationals. A simple sketch should do the trick most of the time with a few points written in. Don't forget your asymptotes and intercepts too!
     
    Last edited: Nov 6, 2009
  5. Nov 6, 2009 #4
    Lastly, your f(5) and f(8) values ARE WRONG. Make sure that you plugged 5 and 8 into f(x) and nothing else. Also, upon graphing, you will note that f(8) is not an absolute max or min. Compare f(0), f(5), and f(8)
     
  6. Nov 6, 2009 #5
    f(5)=1/10
    but i think f(8) is right
     
  7. Nov 6, 2009 #6
    shouldn't that mean that f(0) is a min and f(5) is a max?
     
  8. Nov 6, 2009 #7
    got it. the answer is correct thanks!! :)
     
  9. Nov 6, 2009 #8
    My bad, yes f(8) is right.
     
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