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Math Help - Difference Quotient...HELP!!!

  1. #1
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    Exclamation Difference Quotient...HELP!!!

    This is the problem that I have been struggling with the whole weekend:

    Find the difference quotient of:
    g(x) = (x+1) / (x-3)

    I have:
    g(x) = [ ( (x+h)+1/(x+h)-3 ) - (x+1)/(x-3) ] / h

    **Problem:
    Is the LCD
    1) (x-3)(x+h)-3 or
    2) (x-3)(x+h-3)

    Because grouping changes the whole answer...also, if it's choice 1, is the (-3) at the end of the LCD multiplied times the numerators as well or is it just added after you multiply the rest of the LCD??
    Help!!!
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  2. #2
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    Re: Difference Quotient...HELP!!!

    For your difference quotient, you should have:

    \frac{g(x+h)-g(x)}{h}=\frac{\frac{(x+h)+1}{(x+h)-3}-\frac{x+1}{x-3}}{h}

    Now, what do you think the LCD in the numerator of the difference quotient is?
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  3. #3
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    Re: Difference Quotient...HELP!!!

    LCD is (x-3)(x+h)-3 correct?
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    Re: Difference Quotient...HELP!!!

    So when I multiply by the LCD to cancel out the small fractions, do i multiply it as:
    (x-3)(x+h-3) or (x-3)(x+h)-3 ?
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  5. #5
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    Re: Difference Quotient...HELP!!!

    Use the first choice. Do you see that it has to be the product of the two denominators?
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  6. #6
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    Re: Difference Quotient...HELP!!!

    That's what I chose but I had a feeling a was doing it wrong.
    My final answer is -4
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  7. #7
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    Re: Difference Quotient...HELP!!!

    That is the correct result for the numerator, after dividing out the factor of h common to the numerator and denominator of the difference quotient. So what is the denominator?
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  8. #8
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    Re: Difference Quotient...HELP!!!

    After multiplying by the LCD and adding/subtracting like terms, I got:
    -4h/h
    I then canceled out the h's to get -4 as my overall final answer
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  9. #9
    MHF Contributor MarkFL's Avatar
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    Re: Difference Quotient...HELP!!!

    You must also multiply the denominator of the difference quotient by the LCD, so that in effect you are multiplying the quotient by 1 in the form of LCD/LCD. Otherwise, you are changing the expression. Recall the multiplicative identity 1a = a.
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