Difference Quotient Problem, Please Check My work!

[rpalmier82]

My friend and I are coming up with different approaches for this problem, can someone please tell me which way is the correct way??




EVALUATE THE DIFFERENCE QUOTIENT FOR THE GIVEN FUNCTION
SIMPLIFY YOUR ANSWER:


f(x)= (x+3) / (x+1) ; ( f(x) -f(2) ) / (x-2)


We both had the same initial approach, which was simply to plug x and 2 into their respective places:


[(x+3) / (x-1) - (2+3) / (2-1)] / (x -2)


Which simplifies to:


[(x+3) / (x-1) - 5] / (x-2) <---- I'll refer to this as PART A later


From this point, he multiplied everything by (x-1) to clear the denominator on (x-3) getting:


[(x+3) - 5x+5] / [(x-2)(x-1)] = (-4x+8) / [(x-2)(x-1)] = [-4(x-2)] / (x-2)(x-1) = -4 / (x-1)






In contrast, from Part A, I decided to multiply everything by (x-2) to clear the (x-2) from the problem:


[(x^2+x-6)/(x-1)] - (5x-10) , the multiplied by (x-1)...


x^2 + x - 6 - 5x^2 + 15x -10 = -4x^2 + 16x -16 = -4(x^2+4x-4)




Which is correct??

 

[HallsofIvy]

The first one is correct. You are misunderstanding "multiply everything".
In the first, you have \(\displaystyle \frac{\frac{x+ 3}{x- 1}- 5}{x- 2}\). To "multiply everything by x- 1" here means "multiply both numerator and denominator by x- 1": \(\displaystyle \frac{\frac{x+ 3}{x- 1}(x- 1)- 5(x- 1)}{(x- 1)(x- 2)}\) which does not change the result because you are both multiplying and dividing by x- 1: \(\displaystyle \frac{x- 1}{x- 1}= 1\).

But when you say "multiply everything by x- 2" you are simply multiplying the entire fraction by x- 2. You are not also dividing by x- 2 so you are changing the value.