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Math Help - Calculus Problem

  1. #1
    Junior Member BobBali's Avatar
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    Calculus Problem

    Hi All,

    Question asks to find the slope of tangent from First Principles for f(x) = \sqrt{x-6}  at point of contact x= 10
    First find:

    f(10)= 2

    Then using Slope-Limit equation:  f(x) = \frac{f(x) - f(a)}{x - a}

    \frac{\sqrt{x+10} - \sqrt{6} - 2}{x - 10}

    \frac{\sqrt{x+10} - \sqrt{6} -2}{x - 10}

    \frac{\sqrt{x+10} - \sqrt{6} -2}{\sqrt{x} + \sqrt{10} - \sqrt{x} - \sqrt{-10}

    After this i am at a loss...
    Last edited by BobBali; March 12th 2012 at 01:30 AM.
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  2. #2
    Senior Member
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    Crna Gora
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    Re: Calculus Problem

    Quote Originally Posted by BobBali View Post
    Hi All,

    Question asks to find the slope of tangent from First Principles for f(x) = \sqrt{x-6}  at point of contact x= 10
    First find:

    f(10)= 2

    Then using Slope-Limit equation:  f(x) = \frac{f(x) - f(a)}{x - a}

    \frac{\sqrt{x+10} - \sqrt{6} - 2}{x - 10}

    \frac{\sqrt{x+10} - \sqrt{6} -2}{x - 10}

    \frac{\sqrt{x+10} - \sqrt{6} -2}{\sqrt{x} + \sqrt{10} - \sqrt{x} - \sqrt{-10}

    After this i am at a loss...
    m= \displaystyle 	\lim_{x \to 10} \frac{f(x)-f(a)}{x-a}

    m= \displaystyle 	\lim_{x \to 10} \frac{\sqrt{x-6}-2}{x-10} \cdot \frac {\sqrt{x-6}+2}{\sqrt{x-6}+2}

    m=\displaystyle 	\lim_{x \to 10} \frac{x-10}{(x-10)(\sqrt{x-6}+2)}=\displaystyle 	\lim_{x \to 10} \frac{1}{(\sqrt{x-6}+2)}=\frac{1}{4}
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