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Math Help - Stuck on simple calculus problem...

  1. #1
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    Stuck on simple calculus problem...

    Let


    Defined in the interval 1<(or equal to) x <(or equal to) 9

    I've found the derivative:


    But don't know what to do next....
    Attached Thumbnails Attached Thumbnails Stuck on simple calculus problem...-screen-shot-2010-11-27-6.58.00-pm.png   Stuck on simple calculus problem...-screen-shot-2010-11-27-6.58.13-pm.png  
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  2. #2
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    What exactly is the question?
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  3. #3
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    Oh sorry. The maximum and minimum value (both the x position and the actual value, though obviously finding one it will be easy to find the other). That's why i took the derivative but I don't know how I would set that to zero or what...
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  4. #4
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    Establish same common denominators:

    \displaystyle \frac{(15-x)(2x-6)-2(x^2-6x+25)}{2\sqrt{x^2-6x+25}}=0\rightarrow  (15-x)(2x-6)-2(x^2-6x+25)=0

    and

    whenever the denominator is zero you will have an asymptote unless it factors out and then you have a whole in the graph.

    Your critical values to test are when the numerator and denominator are 0.
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  5. #5
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    \displaystyle \frac{(15-x)(x-3)}{\sqrt{x^2-6x+25}} - \sqrt{x^2-6x+25} = 0

    \displaystyle \frac{(15-x)(x-3)}{\sqrt{x^2-6x+25}} = \sqrt{x^2-6x+25}<br />

    (15-x)(x-3) = x^2-6x+25

    -x^2+18x-45 = x^2-6x+25

    0 = 2x^2-24x+70

    0 = x^2-12x+35

    0 = (x-7)(x-5)

    check the critical values for extrema ... don't forget the endpoints.
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  6. #6
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    Thank you both of you!
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  7. #7
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    Quote Originally Posted by skeeter View Post
    \displaystyle \frac{(15-x)(x-3)}{\sqrt{x^2-6x+25}} - \sqrt{x^2-6x+25} = 0

    \displaystyle \frac{(15-x)(x-3)}{\sqrt{x^2-6x+25}} = \sqrt{x^2-6x+25}<br />

    (15-x)(x-3) = x^2-6x+25

    -x^2+18x-45 = x^2-6x+25

    0 = 2x^2-24x+70

    0 = x^2-12x+35

    0 = (x-7)(x-5)

    check the critical values for extrema ... don't forget the endpoints.
    Oops, I see you factored it out
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