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Thread: evaluating difference quotients

  1. #1
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    evaluating difference quotients

    The problem is $\displaystyle f(x)=1/x$ and the difference quotient to evaluate is $\displaystyle (f(x)-f(a))/x-a$ the answer in the book is -1/ax. How did they arrive at that conclusion?

    Thanks.
    Evan
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    show what have you done.
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    Quote Originally Posted by Evan.Kimia View Post
    The problem is $\displaystyle f(x)=1/x$ and the difference quotient to evaluate is $\displaystyle (f(x)-f(a))/x-a$ the answer in the book is -1/ax. How did they arrive at that conclusion?
    start with...
    $\displaystyle \frac{\frac{1}{x}-\frac{1}{a}}{x-a}$
    getting a common denominator in the numerator...
    $\displaystyle \frac{\frac{a-x}{ax}}{x-a}$
    factor a negative out of the denominator...
    $\displaystyle \frac{\frac{a-x}{ax}}{-(a-x)}$
    cancel the (a-x) and rewrite...
    $\displaystyle \frac{-1}{ax}$
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    i plugged in f(x) and f(a) into the equation and got...

    $\displaystyle (1/x)-(1/a)/(x-a)$

    im not sure how to simplify this numerator correctly.
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    Quote Originally Posted by pflo View Post
    start with...
    $\displaystyle \frac{\frac{1}{x}-\frac{1}{a}}{x-a}$
    getting a common denominator in the numerator...
    $\displaystyle \frac{\frac{a-x}{ax}}{x-a}$
    factor a negative out of the denominator...
    $\displaystyle \frac{\frac{a-x}{ax}}{-(a-x)}$
    cancel the (a-x) and rewrite...
    $\displaystyle \frac{-1}{ax}$
    I see that you factored out a-x by -1 in order to cancel out the a-x above, but what method would i use to get a common denominator in the numerator?

    Thanks.
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  6. #6
    Member pflo's Avatar
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    Quote Originally Posted by Evan.Kimia View Post
    I see that you factored out a-x by -1 in order to cancel out the a-x above, but what method would i use to get a common denominator in the numerator?
    The numerator is $\displaystyle \frac{1}{x}-\frac{1}{a}$.

    Think about how you would do this: $\displaystyle \frac{1}{2}-\frac{1}{3}$.

    To get a common denominator you would multiply the $\displaystyle \frac{1}{2}$ by $\displaystyle \frac{3}{3}$ and the $\displaystyle \frac{1}{3}$ by $\displaystyle \frac{2}{2}$, right?

    $\displaystyle \frac{1}{2}-\frac{1}{3} = \frac{1*3}{2*3}-\frac{1*2}{3*2} = \frac{3-2}{6} = \frac{1}{6}$

    Same thing here using x and a instead of 2 and 3.

    $\displaystyle \frac{1}{x}-\frac{1}{a} = \frac{1*a}{x*a}-\frac{1*x}{a*x} = \frac{a-x}{ax}$
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    Thank you so much everyone.
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