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Finding A Tangent Line(s)

  1. Jun 6, 2012 #1
    Find equations of the tangent lines to the graph of [itex]f(x)=\frac{x}{x-1}[/itex] that pass through the point [itex](-1, 5)[/itex].

    Well, first I took the derivative, and afterwards, I made the connection that the derivative was a slope at any instant on the graph. By this, I inferred that [itex]f'(x) = m[/itex]. I knew that the given point was not on the graph, so I had to find a point on the graph that had the same slope at that point, and also had to pass through the point [itex](-1, 5)[/itex]. So, I calculated the slope to be [itex]m = \frac{y - 5}{x + 1}[/itex]; and so, I set this equal to the derivative, and by doing this I had a system of two equations--those being the derivative and the original function.

    But what followed was great dismay: I found that when I substituted in for y and solved for x, I procured imaginary solutions. I rather thought I was going on a course of brilliance. What did I do wrong?
     
    Last edited: Jun 6, 2012
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  3. Jun 6, 2012 #2

    SammyS

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    Take your equation for the slope, [itex]\displaystyle m = \frac{y - 5}{x + 1}[/itex] and substitute [itex]\displaystyle \frac{x}{x-1}[/itex] in for y, and whatever expression you have for [itex]f'(x)[/itex] in for m . Solve for x to find what location(s) on the graph give the second point for each line.
     
  4. Jun 6, 2012 #3

    HallsofIvy

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    You don't say what you got for the derivative of the curve so it's impossible to tell what, if anything, you did wrong. When I do essentially what you said, I get a quadratic equation for the x coordinate of the point of tangency but it has real roots.
     
  5. Jun 7, 2012 #4
    For my derivative, I get [itex]f'(x) = \frac{-1}{(x-1)^2}[/itex]
     
  6. Jun 7, 2012 #5

    SammyS

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    Good !
     
  7. Jun 8, 2012 #6
    Well, I repeated the process of substituting the the equation [itex]m = \frac{y - 5}{x + 1}[/itex] in for [itex]f'(x)[/itex], which gave [itex]\frac{y - 5}{x + 1} = \frac{-1}{(x - 1)^2}[/itex]. Solving for y gave me [itex]y = \frac{-(x + 1)}{(x - 1)^2} + 5[/itex]. Finally I substituted this into my original function, [itex]\frac{-(x + 1)}{(x - 1)^2} + 5 = \frac{x}{x - 1}[/itex], and solved for x, which gave me [itex]x = \pm2[/itex]. These aren't the correct solutions, though; when I went to graph the function and the two tangents lines, one of them clearly was not a tangent line.

    I attached the graph with the supposed tangent lines.
     

    Attached Files:

  8. Jun 8, 2012 #7
    1. Take your derivative and plug in the x-coordinate you're looking for, this will give you the slope of the tangent (m) at that point x.

    2. Use the point-slope formula to find the tangent line. y-y1=m(x-x1)

    That's all there is to it. I don't know what the deal is with all that substitution and whatnot above, it seems like alot more work than necessary. Maybe I'm misreading something.
     
    Last edited: Jun 8, 2012
  9. Jun 8, 2012 #8

    SammyS

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    Check your solutions to [itex]\displaystyle \frac{-(x + 1)}{(x - 1)^2} + 5 = \frac{x}{x - 1}[/itex] again. x = 2 is correct. x = -2 is not.
     
  10. Jun 9, 2012 #9
    The reason for those seemingly unnecessary steps is that the given coordinate is not a point on the graph of the original function. If you plug -1 into the original function, you'll get (-1, 1/2)

    I got 1/2, would that happened to be the other correct solution?
     
  11. Jun 9, 2012 #10
    Yep. 1/2 and 2 are correct :smile:
     
  12. Jun 9, 2012 #11
    That was a very entertaining problem. Thank you everyone for your insight you so generously provided.
     
  13. Jun 9, 2012 #12
    I see now, I did misread the problem.
     
  14. Jun 9, 2012 #13

    SammyS

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    Here's a graph from WolframAlpha:

    attachment.php?attachmentid=48180&stc=1&d=1339282439.gif
     

    Attached Files:

  15. Jun 9, 2012 #14
    Thank you for the graph, but how were you able to graph multiple equations on one coordinate plane? I've only ever been able to do one at a time.
     
  16. Jun 9, 2012 #15

    SammyS

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    Here's a link to WolframAlpha with the parameters I used.
    http://www.wolframalpha.com/input/?i=plot+x%2F%28x-1%29%2C-x%2B4%2C-4%28x-1%2F2%29-1

    If you don't include the "plot" instruction, but just enter y=x/(x-1),y=-x+4,y=-4(x-1/2)-1, you get this: http://www.wolframalpha.com/input/?i=y%3Dx%2F%28x-1%29%2Cy%3D-x%2B4%2Cy%3D-4%28x-1%2F2%29-1
     
  17. Jun 9, 2012 #16
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