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Math Help - Derivative and Tangent Line Help

  1. #1
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    Post [Solved]Derivative Help

    I need help with these two problems:

    1. Find the Derivative of f(x)=((2*x)/(x=1))^2

    2. Find the points on the graph of the function f(x) = (x+1)/(x-1) where the tangent line(s) to the graph are parallel to the line x+2y = 6.


    Can someone walk me through the second one more carefully? I can't find an example of something like it in my notebook. Thanks for any help in advance.
    Last edited by ebonyscythe; October 4th 2007 at 05:17 AM. Reason: Questions Solved
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  2. #2
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    Hello, ebonyscythe!

    2. Find the points on the graph of the function f(x) \:= \:\frac{x+1}{x-1}
    where the tangent line(s) to the graph are parallel to the line x+2y \:=\: 6

    The line is: . y \:=\:-\frac{1}{2}x+3. . It has slope: m = -\frac{1}{2}


    The slope of the tangent to f(x) is:
    . f'(x) \;=\;\frac{(x-1)\!\cdot\!1 - (x+1)\!\cdot\!1}{(x-1)^2} \;=\;\frac{-2}{(x-1)^2}

    Since they are parallel, their slopes are equal: . -\frac{1}{2}\;=\;\frac{-2}{(x-1)^2}

    Multiply by -2(x-1)^2\!:\;\;(x-1)^2 \:=\:4\quad\Rightarrow\quad x-1\:=\:\pm2\quad\Rightarrow\quad x \:=\:1 \pm 2

    Hence:. x \:=\:3,\,-1

    The corresponding y-values are: . y \:=\:2,\,0


    Therefore, the points are: . (3,\,2),\;(-1,\,0)

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  3. #3
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    Perfect! Thank you... I was thinking that the concept was harder then that, but you made it crystal clear. Thanks again!

    When I worked on question 1., I keep getting a different answer then what my Ti-89 gives me. I seem to have misplaced a negative or something. Here's my work:

    f(x) = (2x/(x+1))^2
    f '(x) = 2(2x/(x+1))((2x*1 - 2(x+1))/(x+1)^2) -->(Chain rule and quotient rule)
    f '(x) = 2((2x/(x+1))*(2x-2(x+1)/(x+1)^2))
    f '(x) = 2((4x^2-4x(x+1))/(x+1)^3)
    f '(x) = 2((4x^2-4x^2-4x)/(x+1)^3)
    f '(x) = -8x/(x+1)^3

    And my calculator gives me 8x/(x=1)^3. What did I do wrong?
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  4. #4
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    Never mind! I realized I was using the quotient rule wrong...
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