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Math Help - Tangent line, derivative, (but really just the last step)

  1. #1
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    Tangent line, derivative, (but really just the last step)

    QUESTION SOLVED, HELP RECEIVED

    So the question is: "At which points do the tangent lines through (1,2) touch y=x/x+1?"

    So I've gone through basically everything (fried my brain this one did):

    1. Found derivative using quotient rule; "1/(x+1)^2
    2. Equalled the slope of the tangent line that passes through (1,2) and (x,x/x+1) to the derivative and solved for x
    3. Got the roots x=-1 and x=-2(+or-)sqrt3
    4. Rejected x=-1 because of restrictions

    Now I believe that the next step is to plug in x into the original y to complete the point. So, err, how do I go about doing that?

    Thanks in advance for any and all help.
    Last edited by Toaster; November 16th 2008 at 03:44 PM. Reason: QUESTION SOLVED, HELP RECEIVED
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  2. #2
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    Quote Originally Posted by Toaster View Post
    QUESTION SOLVED, HELP RECEIVED

    So the question is: "At which points do the tangent lines through (1,2) touch y=x/x+1?"

    So I've gone through basically everything (fried my brain this one did):

    1. Found derivative using quotient rule; "1/(x+1)^2
    2. Equalled the slope of the tangent line that passes through (1,2) and (x,x/x+1) to the derivative and solved for x
    3. Got the roots x=-1 and x=-2(+or-)sqrt3
    4. Rejected x=-1 because of restrictions

    Now I believe that the next step is to plug in x into the original y to complete the point. So, err, how do I go about doing that?
    It's unfortunate that you didn't show how you did steps 2 and 3. I get a quadratic equation which does NOT have those roots.
    Thanks in advance for any and all help.
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  3. #3
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    Ah, well, for curiosity's sake, this is how I did those steps:

    [2-(x/x+1)] / (1-x) = 1/(x+1)^2 (Initial Equation)
    [2(x+1)-x] / (1-x)(x+1) = 1/(x+1)^2 (Found Common Denominator for Left side)
    x+2/(1-x)(x+1) = 1/(x+1)^2 (Simplified left numerator)
    (x+2)(x+1)^2 = (1-x)(x+1) (Cross multiplied; I know now that I could have reduced (x+1) in the denominators before cross multiplying which would mean that I wouldn't have ended up with x=-1 as a root...I also could have cross multiplied without simplifying >//<; )
    x^3 + 5x^2 + 5x + 1 = 0 (Simplified and moved all terms to left side)
    (x+1)(x^2 + 4x + 1) = 0 (Factored using remainder and factor theorem)

    I then rejected x=-1 and used the quadratic equation on the trinomial:
    = [-4 (+or-) sqrt (16-4(1)(1)] / 2(1)
    = [-4 (+or-) sqrt (12)]/ 2
    = [-4 (+or-) 2*sqrt (3)]/ 2
    = -2 (+or-) sqrt 3

    And that's that. I hope that was clear and if you see any mistakes in MY calculations please let me know. ^_^
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