1. Tangent Line

Hello,

I'm getting stuck =/

Problem:

$y = x-1/x-2$
$P(3,2)$

I'm using:

$Lim f(x) - f(a) / x - a$

am I setting it up correctly?

$Lim (x-1/x-2) - 2 / (x - 3)$

2. Originally Posted by l flipboi l
Hello,

I'm getting stuck =/

Problem:

$y = x-1/x-2$
$P(3,2)$

I'm using:

$Lim f(x) - f(a) / x - a$

am I setting it up correctly?

$Lim (x-1/x-2) - 2 / (x - 3)$
Looks OK. You just need to finish. (and lern how to associate properly)

3. That's where I need help, i'm getting stuck...

so far I have..

(x-1)(2x-4)/(x-2)/(x-3)

4. Originally Posted by l flipboi l
That's where I need help, i'm getting stuck...

so far I have..

(x-1)(2x-4)/(x-2)/(x-3)
In this case, it is more practical to use the quotient rule than thae alternate form of the derivative.

$f(x)=\frac{x-1}{x-2}$

So,

$f'(x)=\frac{(x-2)-(x-1)}{(x-2)^2}$

$=-\frac{1}{(x-2)^2}$

Then $m_{tan}=f'(3)=-1$.

This implies that the equation to the tangent line is

$y-2=-(x-3)$ or

$y=-x+5$

5. thanks for the help!

6. I was wondering...can you show me how to do it using this formula Lim f(x) - f(a) / x - a

7. 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)
$\frac{\frac{x-1}{x-2}- 2}{x-3}$

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, $f(x)- f(3)= \frac{x-1}{x-2}- \frac{2x-4}{x-2}$ $= \frac{-x+3}{x-2}= -\frac{x-3}{x-2}$ and
$\frac{f(x)- f(3)}{x- 3}= -\frac{x-3}{(x-2)(x-3)}= -\frac{1}{x-2}$ for x not equal to 3.

Now take the limit of $-\frac{1}{x-2}$ as x goes to 3.

8. cool thanks!