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Thread: Graphing a rational function

  1. March 31st 2010, 12:36 PM #1
    Jools
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    Graphing a rational function

    Got another problem...

    Given f(x) = (x+5)/(2x-4)

    Calculate the first and second derivatives, the intervals on which the function is increasing and decreasing, inflection points, and concavity.

    I've calulated the first and second derivatives and come up with this:

    f '(x) = (2x^2 + 4x - 30) / (2x-4)^2

    f ''(x) = 8(x-17) / (2x-4)^3

    I'm pretty sure those are right... but maybe not. In orer to find the intervals of increase and decrease I should find where f '(x) = 0 right? If that's the case I may need a little push to get that figured out. Based on the second derivative, the inflection point should be x=17 right? That doesn't seem right either... I'm all screwed up on this one I think...
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  2. March 31st 2010, 02:41 PM #2
    skeeter
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    Quote Originally Posted by Jools View Post
    Got another problem...

    Given f(x) = (x+5)/(2x-4)

    Calculate the first and second derivatives, the intervals on which the function is increasing and decreasing, inflection points, and concavity.

    I've calulated the first and second derivatives and come up with this:

    f '(x) = (2x^2 + 4x - 30) / (2x-4)^2

    f ''(x) = 8(x-17) / (2x-4)^3

    I'm pretty sure those are right... but maybe not. In orer to find the intervals of increase and decrease I should find where f '(x) = 0 right? If that's the case I may need a little push to get that figured out. Based on the second derivative, the inflection point should be x=17 right? That doesn't seem right either... I'm all screwed up on this one I think...
    f'(x) is incorrect.
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  3. March 31st 2010, 04:19 PM #3
    Jools
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    Quote Originally Posted by skeeter View Post
    f'(x) is incorrect.
    Gotcha... I just checed f '(x) again and see where I made the errror. Does this look better for f '(x)?

    -14 / (2x-4)^2

    Thanks!
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  4. March 31st 2010, 04:27 PM #4
    skeeter
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    Quote Originally Posted by Jools View Post
    Gotcha... I just checed f '(x) again and see where I made the errror. Does this look better for f '(x)?

    28 / (2x-4)^3

    Thanks!
    f(x) = \frac{x+5}{2x-4}

    f(x) = \frac{1}{2} \cdot \frac{x+5}{x-2}

    f'(x) = \frac{1}{2} \cdot \frac{(x-2)(1) - (x+5)(1)}{(x-2)^2}

    f'(x) = \frac{1}{2} \cdot \frac{x-2 - x - 5}{(x-2)^2}

    f'(x) = -\frac{7}{2(x-2)^2}


    f'(x) = -\frac{7}{2}(x-2)^{-2}

    f''(x) = 7(x-2)^{-3} = \frac{7}{(x-2)^3}
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  5. March 31st 2010, 04:38 PM #5
    Jools
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    Quote Originally Posted by skeeter View Post
    f(x) = \frac{x+5}{2x-4}

    f(x) = \frac{1}{2} \cdot \frac{x+5}{x-2}

    f'(x) = \frac{1}{2} \cdot \frac{(x-2)(1) - (x+5)(1)}{(x-2)^2}

    f'(x) = \frac{1}{2} \cdot \frac{x-2 - x - 5}{(x-2)^2}

    f'(x) = -\frac{7}{2(x-2)^2}


    f'(x) = -\frac{7}{2}(x-2)^{-2}

    f''(x) = 7(x-2)^{-3} = \frac{7}{(x-2)^3}

    Here's how I came up with f '(x)... Can you show me where I went wrong?

    f(x) = (x+5) / (2x-4)

    f(x) = (x+5)(2x-4)^-1
    f '(x) = (x+5)' (2x-4)^-1 + (x+5)(2x-4)^-1'
    = (1)(2x-4)^-1 + (x+5)(-1)(2x-4)^-2(2)
    = (2x-4)^-1 - 2(x+5)(2x-4)^-2
    = 1 / (2x-4) - 2(x+5) / (2x-4)^2
    = (2x-4)/(2x-4)^2 - 2(x+5)/2x-4)^2
    = 2x-4 - 2(x+5) / (2x-4)^2
    = 2x-4-2x-10 / (2x-4)^2
    f '(x) = -14 / (2x-4)^2

    Did I go wrong somewhere? It's close to what you have... Only I've gotten -14 as the numerator instead of -7.
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  6. March 31st 2010, 05:10 PM #6
    skeeter
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    it's the same thing, I just simplified a bit more ...

    \frac{-14}{(2x-4)^2} = \frac{-14}{[2(x-2)]^2} = \frac{-14}{4(x-2)^2}= \frac{-7}{2(x-2)^2}
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  7. March 31st 2010, 05:21 PM #7
    Jools
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    Quote Originally Posted by skeeter View Post
    it's the same thing, I just simplified a bit more ...

    \frac{-14}{(2x-4)^2} = \frac{-14}{[2(x-2)]^2} = \frac{-14}{4(x-2)^2}= \frac{-7}{2(x-2)^2}
    I don't have the 4 in the denominator though...
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  8. March 31st 2010, 05:38 PM #8
    skeeter
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    Quote Originally Posted by Jools View Post
    I don't have the 4 in the denominator though...
    and you shouldn't ... look at the last equation again.
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  9. March 31st 2010, 06:10 PM #9
    Jools
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    Quote Originally Posted by skeeter View Post
    and you shouldn't ... look at the last equation again.
    Sorry I'm missing something here...

    I'm getting

    f '(x) = -14 / (2x-4)^2

    You have -14 / [2(x-2)]^2 which is working out to -14 / 4(x-2)^2. Should you not multiply out the 2(x-2) portion first before squaring it?
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  10. April 1st 2010, 04:36 AM #10
    skeeter
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    Quote Originally Posted by Jools View Post
    Sorry I'm missing something here...

    I'm getting

    f '(x) = -14 / (2x-4)^2

    You have -14 / [2(x-2)]^2 which is working out to -14 / 4(x-2)^2. Should you not multiply out the 2(x-2) portion first before squaring it?
    working with the denominator ...

    <br /> (2x-4)^2 = [2(x-2)]^2 = 2^2(x-2)^2 = 4(x-2)^2<br />

    now put that under -14 ...

    <br /> \frac{-14}{4(x-2)^2} = \frac{-7}{2(x-2)^2} = \frac{-14}{(2x-4)^2}<br />
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