# Domain and range

• Apr 13th 2010, 11:11 PM
toprun91
Domain and range
Hi,
I am stuck with finding the domain and range of $\displaystyle g(x)=\frac{x}{x-1}$. I think that the domain is all values except $\displaystyle x=1$. But what about the range?
• Apr 13th 2010, 11:22 PM
Rapha
Hi there.

Quote:

Originally Posted by toprun91
Hi,
I am stuck with finding the domain and range of $\displaystyle g(x)=\frac{x}{x-1}$. I think that the domain is all values except $\displaystyle x=1$.

That's true,

It is okay to write
$\displaystyle D = \mathbb{R} \backslash \{1\}$

Quote:

Originally Posted by toprun91

I suggest $\displaystyle range(g) = \mathbb{R}$

But I am not sure about it, because a long polynomial division leads to

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

So a(x) is a horizontal asymptote, a(x) tends to +1, if $\displaystyle x \to \pm \infty$

So the range could be $\displaystyle range(g) = \mathbb{R} \backslash \{1 \}$ as well
• Apr 14th 2010, 02:09 AM
HallsofIvy
Another way to find range is to "reverse" the fraction- solve for x- so that domain and range are swapped.

If $\displaystyle y= \frac{x}{x-1}$ then [tex]y(x-1)= xy- y= x. Then xy- x= y so x(y- 1)= y. The fraction inverts into $\displaystyle x= \frac{y}{y- 1}$. Now we can see that y can be any number except 1 so that the range of the original fraction is, indeed, R\{1}.