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Math Help - slope of tangent line on a curve?

  1. #1
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    slope of tangent line on a curve?

    Find the slope of the tangent to the curve f(x) = x^3-4x+1 at the point where x=a. This is what I get
    \lim_{x\to a}\frac{f(x)-f(a)}{x-a}
    \lim_{x\to a}\frac{(x^3-4x+1)-(a^3-4a+1)}{x-a}
    \lim_{x\to a}\frac{(x^3-4x+1)+(-a^3+4a-1)}{x-a}
    \lim_{x\to a}\frac{x^3-4x-a^3+4a}{x-a}

    Then I'm stuck from there. Oh and don't use derivatives or that \frac{dy}{dx} stuff because we haven't learned that yet.
    Last edited by yoman360; August 20th 2009 at 09:55 PM.
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  2. #2
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    Quote Originally Posted by yoman360 View Post
    Find the slope of the tangent to the curve f(x) = x^3-4x+1 at the point where x=a. This is what I get
    \lim_{x\to a}\frac{f(x)-f(a)}{x-a}
    \lim_{x\to a}\frac{(x^3-4x+1)-(a^3-4a+1)}{x-a}
    \lim_{x\to a}\frac{(x^3-4x+1)+(-a^3+4a-1)}{x-a}
    \lim_{x\to a}\frac{x^3-4x-a^3+4a}{x-a}

    Then I'm stuck from there. Oh and don't use derivatives or that \frac{dy}{dx} stuff because we haven't learned that yet.
    I take your last term:

    \lim_{x\to a}\frac{x^3-4x-a^3+4a}{x-a} = \lim_{x\to a}\frac{x^3-a^3 - 4x+4a}{x-a}

    {\color{white}\lim_{x\to a}\frac{x^3-4x-a^3+4a}{x-a}} = \lim_{x\to a}\frac{(x^3-a^3) - 4(x-a)}{x-a}

    {\color{white}\lim_{x\to a}\frac{x^3-4x-a^3+4a}{x-a}} = \lim_{x\to a}\frac{((x-a)(x^2+ax+a^2)) - 4(x-a)}{x-a}

    Now cancel (x-a):

    {\color{white}\lim_{x\to a}\frac{x^3-4x-a^3+4a}{x-a}} = \lim_{x\to a}(x^2+ax+a^2 - 4) = \boxed{3a^2-4}
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