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Math Help - Finding the equation of a tanget line(derivative error)

  1. #1
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    Finding the equation of a tanget line(derivative error)

    In order to find the equation of a tanget line, you need the slope.

    the problem I was given is

    f(x) = ((x^3)+4x-1)(x-2) and the point they want me to find it at is (1,-4)
    my work so far has been
    f'(x) = (x-2)((3x^2)+4) + ((x^3) + 4x-1)
    f'(x) = ((3x^3)+4x-(6x^2)-8 + (x^3) +4x -1
    and finally
    f'(x) = (4x^3)+8x-(6x^2)-9
    Would I be correct in plugging two points into my final f'(x) in order to get the slope?

    I thought I was done but, something is off in my logic and I'm not sure what.
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  2. #2
    MHF Contributor harish21's Avatar
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    f'(1) would be the slope(m).

    and the equation of the tangent line can be found by:

    (y-y_1) = m(x-x_1)
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  3. #3
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    however, in the book they gave me the point (1,-4)

    so plugging f'(1) I get -3.

    and if slope is rise/run then doesn't it make sense I should have to plug something else in as well? I tried however, I have not gotten the answer in the back of the book which is y = -3x -1..

    So I am wondering if I perhaps did my derivative wrong?
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  4. #4
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    e^(i*pi)'s Avatar
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    The derivate is the slope

    ie: \dfrac{\text{ rise}}{\text{ run}} = f^{\prime}(x)

    From f(x) I would expand the brackets rather then use the product rule (but both work)

    (x-2)(x^3+4x-1) = (x^4+4x^2-x) + (-2x^3-8x+2) = x^4-2x^3+4x^2-9x+2

    That should be easier to derive

    f'(x) = 4x^3-6x^2+8x-9

    f'(1) = 4-6+8-9 = -3





    edit: the error was in my derivative
    edit2: don't want to make it too easy
    Last edited by e^(i*pi); October 10th 2010 at 12:28 PM. Reason: bingo
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  5. #5
    MHF Contributor harish21's Avatar
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    Why doesn't it make sense? Do you mean that the slope of a line cannot be negative??

    When a line has a positive slope it goes up left to right.

    When a line has a negative slope it goes down left to right.

    You have mentioned the answer at the back of the book is y = -3x-1..

    So your slope is m = -3.

    The point you have been given is (x_{1}, y_{1}) = (1,-4)

    Now use (y-y_1) = m(x-x_1) to find the equation of the tangent line..
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  6. #6
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    So wait, if I have the derivative (which seems to be correct).

    and I am given the point (1,-4) and I plug 1 into the derivative and get (1, -3), that is all I need to do for slope?

    oh x.x duh.. -3 is the slope at 1,-4... now it makes a bit more sense.. lol. totally was not thinking about the fact that by plugging 1 into the derivative it was giving me the slope.. I for some reason was thinking I would have to plug another number like 2 into it and get the answer.. but, I'm mistaken.. laughable so.
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  7. #7
    MHF Contributor harish21's Avatar
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    f'(1)=-3. That is all you needed to do to find the slope!
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