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Math Help - Function Review

  1. #1
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    Function Review

    Hello, I'm trying to do a review of functions and I'm having difficulty on this one:

     f(x) = \frac 1x

    The problem is

    \frac {f(x) - f(a)}{x-a}

    I've tried a few possible ways, getting the denominators the same in the numerator. No luck. Probably missing something simple.

    Can someone point me in the right direction?
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  2. #2
    Senior Member nikhil's Avatar
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    Lightbulb check this out

    f(x)=1/x so
    x=1/f(x)
    denominator = x-a =(1/[f(x)] - 1/[(f(a)])
    =[f(a)-f(x)]/f(a)f(x)
    =-[f(x)-f(a)]/f(x)f(a)
    where f(x)-f(a)=numerator
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  3. #3
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    Hello Impalord
    Quote Originally Posted by Impalord View Post
    Hello, I'm trying to do a review of functions and I'm having difficulty on this one:

     f(x) = \frac 1x

    The problem is

    \frac {f(x) - f(a)}{x-a}

    I've tried a few possible ways, getting the denominators the same in the numerator. No luck. Probably missing something simple.

    Can someone point me in the right direction?
    You don't say what it is you want to do with \frac {f(x) - f(a)}{x-a}, but if we replace f(x) with \frac1x and f(a) with \frac1a, we get:

    \frac {f(x) - f(a)}{x-a}=\frac{\Big(\dfrac1x-\dfrac1a\Big)}{x-a}=\frac{\Big(\dfrac{a-x}{ax}\Big)}{x-a}=\frac{-\Big(\dfrac{x-a}{ax}\Big)}{x-a}=-\frac{1}{ax}

    Does that help?

    Grandad
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  4. #4
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    Quote Originally Posted by Grandad View Post
    Hello ImpalordYou don't say what it is you want to do with \frac {f(x) - f(a)}{x-a}, but if we replace f(x) with \frac1x and f(a) with \frac1a, we get:

    \frac {f(x) - f(a)}{x-a}=\frac{\Big(\dfrac1x-\dfrac1a\Big)}{x-a}=\frac{\Big(\dfrac{a-x}{ax}\Big)}{x-a}=\frac{-\Big(\dfrac{x-a}{ax}\Big)}{x-a}=-\frac{1}{ax}

    Does that help?

    Grandad
    Yes, thank you very much. For some reason I wasn't seeing how to turn the

    \frac {a-x}{ax} into \frac {-(x-a)}{ax}

    But it's clear now. Thank you very much!
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