# Thread: Help for domain and range

1. ## Help for domain and range

Hi all,

I am having trouble with the process for this problem.

What is the domain and range for this function?
$f(x)=\sqrt{\frac{x-1}{x+1}}
$

Thanks.

2. Originally Posted by romjke
Hi all,

I am having trouble with the process for this problem.

What is the domain and range for this function?
$f(x)=\sqrt{\frac{x-1}{x+1}}
$

Thanks.
Hi

The domain is the set of x for which $f(x)=\sqrt{\frac{x-1}{x+1}}$ exists

f(x) exists when $\frac{x-1}{x+1} \geq 0$

For $x < -1$
$x-1 < 0$ and $x+1 < 0$ therefore $\frac{x-1}{x+1} > 0$

For x=-1 x+1 =0 therefore f(x) is not defined

For $-1 < x < 1$
$x-1 < 0$ and $x+1 > 0$ therefore $\frac{x-1}{x+1} < 0$

For $x \geq 1$
$x-1 \geq 0$ and $x+1 > 0$ therefore $\frac{x-1}{x+1} \geq 0$

The domain is $)-\infty ; -1( \cup [1 ; +\infty($

3. Thank you, but what about the range?

4. Originally Posted by romjke
Thank you, but what about the range?
Calculate the derivative
Find the variations of f
Find the limits at the boundaries of the domain
You will find the range : $[0 ; 1( \cup )1 ; +\infty($