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Math Help - [SOLVED] Slope of a Tangent

  1. #1
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    [SOLVED] Slope of a Tangent

    So.. Math is frustrating =(

    Here is my question:


    Determine the slope of the tangent to y= 1/1-x at the point (2, -1) without constructing a table. Explain what you are doing.



    The related example in the book that they give me for this question was a bit confusing because their point was (4,4).

    With the same question (different function) and the point (4,4) they said:

    To find the equation of the tangent, approximate the slope by setting a value for x close to be, so use 4.001. The slope can be found by using
    m= ( f(4.001) - f(4))/ (4.001 - 4)

    Would have been nice if they had used two different digits for the points! I tried this, and I did not get the right answer.. A forum member helped me with this question before, and they said that my slope was 1, but I end up with a slope of 1999.9

    Very Frustrating! =( any help? I would like to answer the question the way that they do in the book.
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  2. #2
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    Quote Originally Posted by Slipery View Post
    So.. Math is frustrating =(

    Here is my question:


    Determine the slope of the tangent to y= 1/1-x at the point (2, -1) without constructing a table. Explain what you are doing.



    The related example in the book that they give me for this question was a bit confusing because their point was (4,4).

    With the same question (different function) and the point (4,4) they said:

    To find the equation of the tangent, approximate the slope by setting a value for x close to be, so use 4.001. The slope can be found by using
    m= ( f(4.001) - f(4))/ (4.001 - 4)

    Would have been nice if they had used two different digits for the points!
    Not really relevant- the fact that f(4)= 4 doesn't show up in that calculation.

    I tried this, and I did not get the right answer.. A forum member helped me with this question before, and they said that my slope was 1, but I end up with a slope of 1999.9

    Very Frustrating! =( any help? I would like to answer the question the way that they do in the book.
    It's hard to say what you did wrong to get 1999.9 when you do not show what you did. Since f(x)= 1/(1-x), you can approximate the slope at x= 2 by [f(2.0001)- f(2)]/[2.0001- 2]. f(2.0001)= 1/(1- 2.0001)= -1/(1.0001)= -0.9999 and f(2)= -1 while 2.0001- 2= 0.0001. The approximation is (-0.9999-(-1))/0.0001= .0001/0.0001= 1 (I have to say "approximately" because I rounded off -1/1.0001).

    To get the exact value (which really is 1) you have to take the limit of [f(2+h)- f(2)]/h as h goes to 0.
    For any nonzero h, f(2+h)= 1/(1- 2-h)= -1/(1+h) so f(2+h)- f(2)= -1/(1+h)- (-1)= -1/(1+h)+ (1+h)/(1+h)= h/(1+h) so [f(2+h)- f(2)]/h= 1/(1+h) and the limit of that as h goes to 0, 1/(1+0)= 1.
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  3. #3
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    Ahhh, *sigh*. I subtracted it by 1 on the top, and I should have subtracted it by -1 =( Those little mistakes make a big difference. Thanks for that though, it probably would have taken me another half an hour to realize my mistake
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