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Math Help - Tangent Line

  1. #1
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    Tangent Line

    Hello,

    Can someone please help me with the algebra?

    I'm getting stuck =/

    Problem:

    y = x-1/x-2
    P(3,2)

    I'm using:

    Lim f(x) - f(a) / x - a

    am I setting it up correctly?

    Lim  (x-1/x-2) - 2 / (x - 3)
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by l flipboi l View Post
    Hello,

    Can someone please help me with the algebra?

    I'm getting stuck =/

    Problem:

    y = x-1/x-2
    P(3,2)

    I'm using:

    Lim f(x) - f(a) / x - a

    am I setting it up correctly?

    Lim (x-1/x-2) - 2 / (x - 3)
    Looks OK. You just need to finish. (and lern how to associate properly)
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  3. #3
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    That's where I need help, i'm getting stuck...

    so far I have..

    (x-1)(2x-4)/(x-2)/(x-3)
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by l flipboi l View Post
    That's where I need help, i'm getting stuck...

    so far I have..

    (x-1)(2x-4)/(x-2)/(x-3)
    In this case, it is more practical to use the quotient rule than thae alternate form of the derivative.

    f(x)=\frac{x-1}{x-2}

    So,

    f'(x)=\frac{(x-2)-(x-1)}{(x-2)^2}

    =-\frac{1}{(x-2)^2}


    Then m_{tan}=f'(3)=-1.

    This implies that the equation to the tangent line is

    y-2=-(x-3) or

    y=-x+5
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  5. #5
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    thanks for the help!
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  6. #6
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    I was wondering...can you show me how to do it using this formula Lim f(x) - f(a) / x - a
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  7. #7
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    First, either use LaTex or use parentheses!
    f(x)= (x-1)/(x-2), f(3)= 2 so you have
    [(x-1)/(x-2)- 2]/(x- 3)
    \frac{\frac{x-1}{x-2}- 2}{x-3}

    For the numerator, we have f(x)- f(3)= (x-1)/(x-2)- 2(x-2)/(x-2)= (x-1- (2x-4))/(x-2)= (-x+3)/(x-2) so the entire difference quotient is
    (f(x)-f(3))/(x- 3)= (-x+3)/((x-2)(x-3))= -1/(x-2) because -x+3= -(x-3) and, for x not equal to 3, the x-3 terms cancel

    In Latex, f(x)- f(3)= \frac{x-1}{x-2}- \frac{2x-4}{x-2} = \frac{-x+3}{x-2}= -\frac{x-3}{x-2} and
    \frac{f(x)- f(3)}{x- 3}= -\frac{x-3}{(x-2)(x-3)}= -\frac{1}{x-2} for x not equal to 3.

    Now take the limit of -\frac{1}{x-2} as x goes to 3.
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  8. #8
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    cool thanks!
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