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Thread: 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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    $\displaystyle f(x) = \frac{2}{(x-1)^2}$

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

    point not on curve $\displaystyle (1,1)$

    slope between points ...

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

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

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

    $\displaystyle 2 - (x-1)^2 = -4$

    $\displaystyle (x-1)^2 = 6$

    $\displaystyle x = 1 \pm \sqrt{6}$

    $\displaystyle y = \frac{1}{3}$ 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?

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

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

    $\displaystyle y=mx+b $

    $\displaystyle b= 1-m $

    $\displaystyle y=f(x)
    f'(x)=m$

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

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

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

    $\displaystyle x= \frac{-2}{5}$

    What am i doing wrong?

    i was following these instructions...:
    http://www.karlscalculus.org/l4_5.html
    Last edited by Saibot; Feb 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?

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

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

    $\displaystyle y=mx+b $

    $\displaystyle b= 1-m $

    $\displaystyle y=f(x)
    f'(x)=m$

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

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


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


    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 $\displaystyle f(x) = \frac{2}{(x-1)^2}$ instead of $\displaystyle f(x) = \frac{2}{(1-x)^2}$
    (pardon my dyslexia)

    however ...

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

    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; Feb 20th 2009 at 06:29 AM.
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