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Math Help - Equations of lines tangent to a curve

  1. #1
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    Equations of lines tangent to a curve

    I'm stuck on the following problem:

    "Find the equations of both lines through the point (1,2) that are tangent to the curve y=x/(x+1)"

    So far, I have that y'=1/(x+1)^2. I tried setting that equal to the slope (2-(x/(x+1)))/(1-x) but it isn't working out. Not sure where to go from here, any help would be much appreciated
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  2. #2
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    Quote Originally Posted by maryanna91 View Post
    I'm stuck on the following problem:

    "Find the equations of both lines through the point (1,2) that are tangent to the curve y=x/(x+1)"

    So far, I have that y'=1/(x+1)^2. I tried setting that equal to the slope (2-(x/(x+1)))/(1-x) but it isn't working out. Not sure where to go from here, any help would be much appreciated

    What you wanted to do is right: to compare the slope between (1,2) and a generic point

    \left(x,\frac{x}{x+1}\right) on the function's graph, and the slope of the tangent

    line at this point given by the derivative:

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

     x_{1,2}=-2\pm\sqrt{3} , and thus the slopes are \displaystyle{m_{1,2}=\frac{1}{4\pm 2\sqrt{3}}=1\pm \frac{\sqrt{3}}{2}

    Now use the well-known formula for the equation of a line through a point given its

    slope to find both equations.

    Tonio
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  3. #3
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    Looks like my algebra was just off, thanks a lot for your help!
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