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Find the tangent lines to the curve

  1. Jul 21, 2012 #1
    1. How many tangent lines to the curve [itex]\left(y=\frac{x}{x + 1}\right)[/itex] pass through the point (1,2)? At which points do these tangent lines touch the curve?



    2. [itex]\frac{x}{x + 1}[/itex]



    3. I tried to use the quotient rule and came up with the equation [itex]\frac{1}{(x + 1)^{2}}[/itex]. I tried plugging in 1 to get the slope of 1/4 and the equation y = 1/4x + 7/4. I know that there are two lines, because of the square.
     
  2. jcsd
  3. Jul 21, 2012 #2

    tiny-tim

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    hi frosty8688! :smile:
    because 1 is the value of x at (1,2) ?

    noooo :redface:

    (1,2) isn't on the curve, is it? :wink:
     
  4. Jul 21, 2012 #3
    I did figure out the x values -2±[itex]\sqrt{3}[/itex]. I'm just having problems finding the y values.
     
  5. Jul 21, 2012 #4
    Would it be -2±[itex]\sqrt{3}[/itex] + 1 / 2 which is equal to 1/2 * -1±√3
     
  6. Jul 21, 2012 #5

    HallsofIvy

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    Those are the x values of what points? The points of tangency? If so, y is given by x/(x+1).

    Let [itex](x_0, y_0)= (x_0, x_0/(x_0+1))[/itex] be the point of tangency. Then a line through that point and (1, 2) is given by [itex]y= [(y_0- 2)/(x_0-1)](x- 1)+ 2[/itex]. With y= x/(x+1), [itex]y'= 1/(x+1)^2[/itex] so we must have [itex](y_0- 2)/(x_0-1)= 1/(x_0+1)^2[/itex] as well as [itex]y_0= x_0/(x_0+1)[/itex].

    Solve those two equations for [itex]x_0[/itex] and [itex]y_0[/itex].
     
  7. Jul 21, 2012 #6
    Here's how I found the x values; [itex]\frac{1}{(x+1)^{2}}[/itex] = [itex]\frac{y-2}{x-1}[/itex], y-2 = [itex]\frac{x-1}{(x+1)^{2}}[/itex], [itex]\frac{x}{x+1}-2[/itex] = [itex]\frac{x-1}{(x+1)^{2}}[/itex], [itex]\frac{[x-2(x+1)]}{x+1}[/itex]=[itex]\frac{x-1}{(x+1)^{2}}[/itex], [itex]\frac{x-2x-2}{x+1}[/itex] = [itex]\frac{x-1}{(x+1)^{2}}[/itex], [itex]\frac{-(x+2)}{x+1}[/itex] = [itex]\frac{x-1}{(x+1)^{2}}[/itex], -(x+2) = [itex]\frac{x-1}{x+1}[/itex], (x+2)(x+1) = 1-x, x[itex]^{2}[/itex]+4x+1 = 0. Then I used the quadratic formula and got the x values to be -2±[itex]\sqrt{3}[/itex]
     
  8. Jul 21, 2012 #7
    For the y values, my calculator is giving me the answer [itex]\frac{1±\sqrt{3}}{2}[/itex] and I am wondering how do I get that answer without using the calculator.
     
  9. Jul 21, 2012 #8
    Do I have to multiply [itex]\frac{-2±\sqrt{3}}{-2±\sqrt{3}+1}[/itex] by something.
     
  10. Jul 21, 2012 #9
    I figured it out.
     
  11. Jul 21, 2012 #10
    Even though both have equal gradient, the 2 lines might be of parallel tangent lines.
    Reference to a point on the curve for gradient will ensure both are intersecting same points.
     
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