# Difference Quotient...HELP!!!

• Nov 25th 2012, 09:28 PM
rrooggeerr
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!!!
• Nov 25th 2012, 09:35 PM
MarkFL
Re: Difference Quotient...HELP!!!
For your difference quotient, you should have:

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

Now, what do you think the LCD in the numerator of the difference quotient is?
• Nov 25th 2012, 09:37 PM
rrooggeerr
Re: Difference Quotient...HELP!!!
LCD is (x-3)(x+h)-3 correct?
• Nov 25th 2012, 09:40 PM
rrooggeerr
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 ?
• Nov 25th 2012, 10:26 PM
MarkFL
Re: Difference Quotient...HELP!!!
Use the first choice. Do you see that it has to be the product of the two denominators?
• Nov 25th 2012, 10:29 PM
rrooggeerr
Re: Difference Quotient...HELP!!!
That's what I chose but I had a feeling a was doing it wrong.
• Nov 25th 2012, 10:37 PM
MarkFL
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?
• Nov 25th 2012, 10:40 PM
rrooggeerr
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
• Nov 25th 2012, 10:45 PM
MarkFL
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 1·a = a.