1. ## domain/range help

Hey guys, I'm having a problem with finding the domain and range of this function:

f(x) = sqrt(X^2 - 1)

this is what i thought:
x|xER
y|yER, y≥1

but when you substitute 1 for x, you get y = 0, so it disproves what I thought to be correct.

the second one I need help with is the function:

f(x) = sqrt(x^2 - 1)/x-1

i have no idea what the domain/range for that function could be, i've never seen something like that.

2. $\displaystyle f(x) = \sqrt{x^2 - 1}$

In order for the square root to be defined, $\displaystyle x^2 - 1$ must be a nonnegative number. Therefore, the domain is all x such that $\displaystyle x^2 - 1 \geq 0$ (can you put this in terms of intervals?). The range is $\displaystyle y \geq 0$ (why?).

3. hi
the range is $\displaystyle [0,\infty)$.

4. wow thanks for the fast response.

can
$\displaystyle x^2 - 1 \geq 0$

be rephrased as

x ≥ sqrt(1) ?

thanks again

5. Originally Posted by snypeshow
wow thanks for the fast response.

can
$\displaystyle x^2 - 1 \geq 0$

be rephrased as

x ≥ sqrt(1) ?

thanks again
Not exactly. You have $\displaystyle x^2 \geq 1$, which leads to the inequality $\displaystyle |x| \geq 1$. In summary: x can be negative.

6. so would it be best to leave it as
$\displaystyle x^2 - 1 \geq 0$?

i also need help with:

f(x) = sqrt(x^2 - 1)/x-1

for domain and range

7. Originally Posted by snypeshow
so would it be best to leave it as
$\displaystyle x^2 - 1 \geq 0$?
The best way to write it would be $\displaystyle x \leq -1$ or $\displaystyle x \geq 1$.