# Range of function

• Jan 17th 2010, 02:36 PM
Tweety
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,
• Jan 17th 2010, 02:51 PM
VonNemo19
Quote:

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 $\displaystyle y$
• Jan 17th 2010, 03:11 PM
Tweety
Quote:

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'?
• Jan 17th 2010, 03:13 PM
VonNemo19
Quote:

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.
• Jan 17th 2010, 03:26 PM
Tweety
Quote:

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.
$\displaystyle y = \frac{2x-3}{x-2}$
• Jan 17th 2010, 03:33 PM
VonNemo19
Quote:

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.

• Jan 18th 2010, 04:00 AM
Tweety
Quote:

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?
• Jan 18th 2010, 04:42 AM
Stroodle
Quote:

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 $\displaystyle x>2$
• Jan 18th 2010, 04:44 AM
Raoh
Quote:

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 $\displaystyle (2,\infty )$.
• Jan 18th 2010, 05:22 AM
Raoh
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).