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Math Help - Finding tangent of F(x) if we know tangent passes through a point f(x) does not...

  1. #1
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    Finding tangent of F(x) if we know tangent passes through a point f(x) does not...

    I cant see to understand how to do this question.

    f(x) = 2/(1-x)^2

    Find tangent if tangent goes through (1,1)
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  2. #2
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    f(x) = \frac{2}{(x-1)^2}

    point on curve ... \left(x , \frac{2}{(x-1)^2}\right)

    point not on curve (1,1)

    slope between points ...

    \frac{\frac{2}{(x-1)^2} - 1}{x-1} = \frac{2}{(x-1)^3} - \frac{1}{x-1} = \frac{2 - (x-1)^2}{(x-1)^3}

    f'(x) = -\frac{4}{(x-1)^3}

    \frac{2 - (x-1)^2}{(x-1)^3} = -\frac{4}{(x-1)^3}<br />

    2 - (x-1)^2 = -4

    (x-1)^2 = 6

    x = 1 \pm \sqrt{6}

    y = \frac{1}{3} for both x-values

    note that two tangent lines exist with the desired properties.

    finish up by finding the tangent line equations
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    Actually i did it this way and i got something totally different. What am i doing wrong?

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

    f'(x)=\frac{-4}{(1-x)^3}

     y=mx+b

     b= 1-m

     y=f(x)<br />
f'(x)=m

    \frac{2}{(1-x)^2}=(\frac{-4}{(1-x)^3})x + 1 + \frac{4}{(1-x)^3}

    \frac{2(1-x)-4x+x-4}{(1-x)^3}=0

    \frac{-2-5x}{(1-x)^3} = 0

    x= \frac{-2}{5}

    What am i doing wrong?

    i was following these instructions...:
    http://www.karlscalculus.org/l4_5.html
    Last edited by Saibot; February 19th 2009 at 07:32 PM.
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  4. #4
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    Quote Originally Posted by Saibot View Post
    Actually i did it this way and i got something totally different. What am i doing wrong?

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

    f'(x)=\frac{-4}{(1-x)^3}

     y=mx+b

     b= 1-m

     y=f(x)<br />
f'(x)=m

    \frac{2}{(1-x)^2}=(\frac{-4}{(1-x)^3})x + 1 + \frac{4}{(1-x)^3} this step is fine.

    \frac{2(1-x)-4x+x-4}{(1-x)^3}=0 this step has errors. should start out as ...


    \frac{2}{(1-x)^2}=\frac{-4x}{(1-x)^3} + \textcolor{red}{\frac{(1-x)^3}{(1-x)^3}} + \frac{4}{(1-x)^3}


    i was following these instructions...:
    Karl's Calculus Tutor - Alternative Solution to Tangent Problem
    first, I'll admit to a mistake ... I "saw" the original function as f(x) = \frac{2}{(x-1)^2} instead of f(x) = \frac{2}{(1-x)^2}
    (pardon my dyslexia)

    however ...

    \frac{2}{(x-1)^2} = \frac{2}{(1-x)^2}

    so, my solution should be the same. I'll have to work on it and try to get back to you.
    Last edited by skeeter; February 20th 2009 at 06:29 AM.
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