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F(x)=2x + 32/x find the absolute maximum and minimum if they exist

  1. May 8, 2009 #1
    find the absolute maximum and absolute minimum (if any) for the given function :
    f(x)= 2x + 32/x ; when x > 0


    I tried to solve it but i'm not sure if its a maxima or minima, here are my try :
    f(x)= 2x + 32/x
    f'(x)= 2-32/x^2
    2-32/x^2=0
    2u^2=32
    u^2=16
    u=4 or - 4
    -4 is not included cause x > 0

    f(4)= 2(4) + 32/4
    f(4)= 8 + 8
    f(4)= 16
    (4, 16) but i'm not sure if it an absolute maximum or minimum?

    are my steps correct ?
     
  2. jcsd
  3. May 8, 2009 #2

    dx

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    Find the second derivative at x = 4 to determine if it's a minimum or a maximum. If f''(4) > 0, then it's a local minimum, and if f''(4) < 0 it's a local maximum. Since the only x satisfying x > 0 and f'(x) = 0 is at x = 4, you can tell that it must be an absolute extremum.
     
  4. May 8, 2009 #3
    thank you
    but isn't the second derivative test only to determine the max or min relative ? not absolute extrema?
     
  5. May 8, 2009 #4

    dx

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    Yes, but if you find only one local maximum and no other places where f'(x) = 0, it must be an absolute maximum. Try drawing a graph. Draw a local maximum, and try to continue the curve and make it attain another higher maximum. Can you get it to do that without creating a local minimum somewhere?
     
  6. May 8, 2009 #5
    thanks for your help :)
     
  7. May 9, 2009 #6
    i got another quistion
    do i have to find f(-4) and f''(-4) ??? or is it just f(4) and f''(-4) ?
    cause it is > o
     
  8. May 9, 2009 #7

    Mark44

    Staff: Mentor

    The problem as given asked for the maximum and minimum of your function for x > 0, so no, you don't need to check x = -4.
     
  9. May 9, 2009 #8

    jbunniii

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    Let's take a step back and recall something important.

    A function is guaranteed to have an absolute maximum and an absolute minimum on a closed and bounded (aka compact) interval [a,b] if it is continuous everywhere in that interval.

    Now, your function is continuous for all x > 0, but you are looking for a maximum and minimum on the interval (0, infinity), and this interval is NOT compact. Therefore the function is NOT guaranteed to have either a maximum or a minimum on that interval!

    In fact, it certainly has no absolute maximum: if you plug in increasingly smaller numbers for x, then 2x gets closer and closer to 0, but 32/x gets larger and larger without bound! Similarly, if you pick x larger and larger, 32/x gets closer and closer to zero, but 2x grows without bound. Thus I could make f(x) equal a trillion or a googol or any big number I like if I choose a small enough x or a large enough x.

    So at best you can hope to find an absolute minimum.

    To verify that an absolute maximum does exist, let's pick x1 and x2 so that

    [tex]f(x1) \geq 1000000000000[/tex] for [tex]0 \leq x < x1[/tex] or [tex]x > x2[/tex]

    Then [tex][x1,x2][/tex] is a compact interval, and f is continuous at every point in that interval, so it has an absolute minimum (and absolute maximum) when restricted to that interval. As long as the absolute minimum on [tex][x1,x2][/tex] is less than 1000000000000, then it is also the absolute minimum on all of [tex](0,\infty)[/tex].

    Since f is also differentiable on all of [tex][x1,x2][/tex], you can now proceed with the usual first and second derivative tests (along with checking the endpoints) to find the absolute max and min on [tex][x1,x2][/tex].
     
    Last edited: May 9, 2009
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