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Absolute minimum and maximum

  1. Mar 18, 2009 #1
    1. The problem statement, all variables and given/known data
    FIND THE ABSOLUTE MINIMUM AND ABSOLUTE MAXIMUM OF:

    f(x) = 9x + 1/x
    on the interval [1,3]

    2. Relevant equations



    3. The attempt at a solution

    I dont know how to get started!!!
     
  2. jcsd
  3. Mar 18, 2009 #2

    lanedance

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    Homework Helper

    hi silverbomb20 - any ideas or thoughts on what defines a maxima or minima?
     
  4. Mar 18, 2009 #3
    i know i need to use the power rule to get it started, but i dont know how to apply it to the fraction..


    derivative of x ^n is n * x^(n-1) right?

    i know the 9x would go to just 9
    but i am unsure about the fraction
     
  5. Mar 18, 2009 #4

    Mark44

    Staff: Mentor

    Yes, d/dx(xn) = nxn-1. And 1/x = x-1, so you can use the power rule on that.
     
  6. Mar 18, 2009 #5

    lanedance

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    first part sounds good

    so we're stuck on
    [tex] \frac{d}{dx} (\frac{1}{x}) [/tex]

    you could write it like below and use the power rule
    [tex] \frac{d}{dx} (x^{-1}) [/tex]

    or you could try and do it from first principles...
     
  7. Mar 18, 2009 #6
    okay so would it be 9 + -1x^-2 ?

    im a history major having a lot of trouble with calculus...please help me
     
  8. Mar 18, 2009 #7

    lanedance

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    yes, that looks correct

    so when does

    [tex] 9-\frac{1}{x^2} = 0 [/tex] ?
     
  9. Mar 19, 2009 #8
    in my mind...never.

    f1(x) = 9 + -1x^-2

    so i have to set it equal to zero right?

    so 9 + -1x^-2=0

    but that doesnt factor
     
  10. Mar 19, 2009 #9

    Mark44

    Staff: Mentor

    Multiply both sides by x^2.

    Before we get (more) bogged down with minutiae, let's look at the strategy. Extreme values of a function f will happen at values of x for which f'(x) is zero, or at endpoints of the interval of definition.
     
  11. Mar 19, 2009 #10

    lanedance

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    f'(x) represents a critical point which could be a maxima, minima or point of inflection, but you nee dto check to find which

    as we are talking about absolute max & min we also need to check the boundaries of [1,3] ie the points x=1 and x=3

    as for the local minima you're on your way
    9 + -1x^-2=0

    try first multiplying both sides of the equation by x^2
     
  12. Mar 19, 2009 #11
    okay so thats gonna give me 9x^2=x^2?

    or does that -1x^-2 not cancel?
     
  13. Mar 19, 2009 #12

    Mark44

    Staff: Mentor

    No. Try again.
     
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