1. ## Range of function

Find the range of $\displaystyle \frac{2x-3}{x-2}$
$\displaystyle x> 2$ x ∈ R

I keep getting stuck on questions about finding the range, can someone please show me if there is a general rule? And how would I go about finding the range of this function?

Thank you,

2. Originally Posted by Tweety
Find the range of $\displaystyle \frac{2x-3}{x-2}$
$\displaystyle x> 2$ x ∈ R

I keep getting stuck on questions about finding the range, can someone please show me if there is a general rule? And how would I go about finding the range of this function?

Thank you,
Let $\displaystyle y=\frac{2x-3}{x-2}$.

Interchange variables and then solve for $\displaystyle y$

3. Originally Posted by VonNemo19
Let $\displaystyle y=\frac{2x-3}{x-2}$.

Interchange variables and then solve for $\displaystyle y$
Thanks,

Do you mean like this:

$\displaystyle x = \frac{2y-3}{y-2}$

and than solve for 'y'?

4. Originally Posted by Tweety
Thanks,

Do you mean like this:

$\displaystyle x = \frac{2y-3}{y-2}$

and than solve for 'y'?
Yes, and then take note of the implied domain of x. This is the range that you seek.

5. Originally Posted by VonNemo19
Yes, and then take note of the implied domain of x. This is the range that you seek.
Okay but I am quite sure what you mean, when I solve for y I get the same answer.
$\displaystyle y = \frac{2x-3}{x-2}$

6. Originally Posted by Tweety
Okay but I am quite sure what you mean, when I solve for y I get the same answer.
$\displaystyle y = \frac{2x-3}{x-2}$
Damn, tweety. I'm sorry. I was absent minded just then. Don't exchange variables. just solve for x outright.

7. Originally Posted by VonNemo19
Damn, tweety. I'm sorry. I was absent minded just then. Don't exchange variables. just solve for x outright.
Okay thanks,
so when I solve I get $\displaystyle x = \frac{3y-2}{y-2}$ so the range is all the values y can take on, would that mean the range is y>2 ?

And also can I use this method for all functions involving a fraction like this?

8. Originally Posted by Tweety
Okay thanks,
so when I solve I get $\displaystyle x = \frac{3y-2}{y-2}$ so the range is all the values y can take on, would that mean the range is y>2 ?

And also can I use this method for all functions involving a fraction like this?
Another method is to rearrange the equation to make it easier to see the range:

$\displaystyle y=\frac{2x-3}{x-2}$

$\displaystyle y=\frac{2(x-2)+1}{x-2}$

$\displaystyle y=2+\frac{1}{x-2}$

*Edit - Remembering that $\displaystyle x>2$

9. Originally Posted by Tweety
Okay thanks,
so when I solve I get $\displaystyle x = \frac{3y-2}{y-2}$ so the range is all the values y can take on, would that mean the range is y>2 ?

And also can I use this method for all functions involving a fraction like this?
the range is $\displaystyle (2,\infty )$.

10. if we take $\displaystyle y=2+\frac{1}{x-2}$
$\displaystyle \lim_{x\to 2^+}\left (2+\frac{1}{x-2} \right )=\infty$
and,
$\displaystyle \lim_{x\to \infty }\left (2+\frac{1}{x-2} \right )=2$
wouldn't now be the range $\displaystyle (\infty ,2)$.
(i might be wrong though).