Hello,

Can someone please help me with the algebra?

I'm getting stuck =/

Problem:

I'm using:

am I setting it up correctly?

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- Jan 26th 2010, 07:12 PMl flipboi lTangent Line
Hello,

Can someone please help me with the algebra?

I'm getting stuck =/

Problem:

I'm using:

am I setting it up correctly?

- Jan 26th 2010, 07:14 PMVonNemo19
- Jan 26th 2010, 07:22 PMl flipboi l
That's where I need help, i'm getting stuck...

so far I have..

(x-1)(2x-4)/(x-2)/(x-3) - Jan 26th 2010, 07:31 PMVonNemo19
- Jan 27th 2010, 01:26 AMl flipboi l
thanks for the help!

- Jan 27th 2010, 01:42 AMl flipboi l
I was wondering...can you show me how to do it using this formula Lim f(x) - f(a) / x - a

- Jan 27th 2010, 04:05 AMHallsofIvy
First, either use LaTex or use parentheses!

f(x)= (x-1)/(x-2), f(3)= 2 so you have

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

For the numerator, we have f(x)- f(3)= (x-1)/(x-2)- 2(x-2)/(x-2)= (x-1- (2x-4))/(x-2)= (-x+3)/(x-2) so the entire difference quotient is

(f(x)-f(3))/(x- 3)= (-x+3)/((x-2)(x-3))= -1/(x-2) because -x+3= -(x-3) and, for x not equal to 3, the x-3 terms cancel

In Latex, and

for x not equal to 3.

Now take the limit of as x goes to 3. - Jan 27th 2010, 10:46 AMl flipboi l
cool thanks!