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Simple calc question

  1. Nov 3, 2006 #1
    y= x / (x + 1) I need to find the equation of every line tangent to the curve which passs through the point (-1, 3). The problem arises when you find the first derivative then plug in the x value which results in an answer of 0 in the denominator. Any help???
  2. jcsd
  3. Nov 3, 2006 #2


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    You're looking for all sets of points (x, y), s.t. y=mx+b, 3 = -1*m + b,
    y= x / (x + 1) and m = 1/(x+1)^2
  4. Nov 3, 2006 #3


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    You don't plug x= -1 into the equation! Obviously (-1,3) is not on the given curve. You are looking for tangent lines to that curve that pass through the point (-1, 3) off the curve.

    You are looking for the tangent line at (x0,y0) that passes through (-1, 3). If y= x / (x + 1) , then y'= m= 1/(x+1)2. The equation of the line passing through (-1, 3) with slope m is y= m(x+1)+ 3. You are looking for x0 such that
    [tex]y= \frac{1}{(x_0+1)^2}(x+1)+ 3[/tex]
    is passes through [itex](x_0,\frac{x_0}{x_0+1}[/tex]
    That is,
    [tex]\frac{x_0}{x_0+1}= \frac{1}{(x_0+1)^2}(x_0+1)+ 3[/tex]
    Solve that for x0.
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