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Math Help - Dumb tangent line question

  1. #1
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    Dumb tangent line question

    f(x) = x^3 - 3x^2 - x +3

    I need to find the slope of the tangent line at x=2 using lim as x->2
    (f(x) - f(2)) / (x-2)

    I cannot for the life of me factor the equation in order to cancel out the x - 2. I feel pretty dumb, but I'm tired, and I need to do this for a write-up.
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  2. #2
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    f(x) - f(2) =

    (x^3 - 3x^2 - x +3) - (-3) =

    x^3 - 3x^2 - x + 6

    divide the cubic polynomial by (x - 2) using long division or synthetic division, you will find that it factors as ...

    (x-2)(x^2-x-3)

    \lim_{x \to 2} (x^2 - x - 3) = ?
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by alexmcentee View Post
    f(x) = x^3 - 3x^2 - x +3

    I need to find the slope of the tangent line at x=2 using lim as x->2
    (f(x) - f(2)) / (x-2)

    I cannot for the life of me factor the equation in order to cancel out the x - 2. I feel pretty dumb, but I'm tired, and I need to do this for a write-up.
    Using this definition, we see that \lim_{x\to 2}\frac{x^3-3x^2-x+3-(-3)}{x-2}=\lim_{x\to 2}\frac{x^3-3x^2-x+6}{x-2}

    But x^3-3x^2-x+6=(x-2)(x^2-x-3). You can use synthetic division with the root being 2, since x-2=0 => x=2 to get this factorized form.

    Thus, \lim_{x\to 2}\frac{x^3-3x^2-x+6}{x-2}=\lim_{x\to 2}x^2-x-3=\dots

    Does this make sense?

    --Chris
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