Hi all,
I am having trouble with the process for this problem.
What is the domain and range for this function?
$\displaystyle f(x)=\sqrt{\frac{x-1}{x+1}}
$
Thanks.
Hi
The domain is the set of x for which $\displaystyle f(x)=\sqrt{\frac{x-1}{x+1}}$ exists
f(x) exists when $\displaystyle \frac{x-1}{x+1} \geq 0$
For $\displaystyle x < -1$
$\displaystyle x-1 < 0$ and $\displaystyle x+1 < 0$ therefore $\displaystyle \frac{x-1}{x+1} > 0$
For x=-1 x+1 =0 therefore f(x) is not defined
For $\displaystyle -1 < x < 1$
$\displaystyle x-1 < 0$ and $\displaystyle x+1 > 0$ therefore $\displaystyle \frac{x-1}{x+1} < 0$
For $\displaystyle x \geq 1$
$\displaystyle x-1 \geq 0$ and $\displaystyle x+1 > 0$ therefore $\displaystyle \frac{x-1}{x+1} \geq 0$
The domain is $\displaystyle )-\infty ; -1( \cup [1 ; +\infty($