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Math Help - evaluating difference quotients

  1. #1
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    evaluating difference quotients

    The problem is f(x)=1/x and the difference quotient to evaluate is (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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  2. #2
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    show what have you done.
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    Quote Originally Posted by Evan.Kimia View Post
    The problem is f(x)=1/x and the difference quotient to evaluate is (f(x)-f(a))/x-a the answer in the book is -1/ax. How did they arrive at that conclusion?
    start with...
    \frac{\frac{1}{x}-\frac{1}{a}}{x-a}
    getting a common denominator in the numerator...
    \frac{\frac{a-x}{ax}}{x-a}
    factor a negative out of the denominator...
    \frac{\frac{a-x}{ax}}{-(a-x)}
    cancel the (a-x) and rewrite...
    \frac{-1}{ax}
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  4. #4
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    i plugged in f(x) and f(a) into the equation and got...

    (1/x)-(1/a)/(x-a)

    im not sure how to simplify this numerator correctly.
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  5. #5
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    Quote Originally Posted by pflo View Post
    start with...
    \frac{\frac{1}{x}-\frac{1}{a}}{x-a}
    getting a common denominator in the numerator...
    \frac{\frac{a-x}{ax}}{x-a}
    factor a negative out of the denominator...
    \frac{\frac{a-x}{ax}}{-(a-x)}
    cancel the (a-x) and rewrite...
    \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 \frac{1}{x}-\frac{1}{a}.

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

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

    \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.

    \frac{1}{x}-\frac{1}{a} = \frac{1*a}{x*a}-\frac{1*x}{a*x} = \frac{a-x}{ax}
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  7. #7
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    Thank you so much everyone.
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