# Thread: Finding the Domain (really simple)

1. ## Finding the Domain (really simple)

What is the domain of the functions:

g(x)= square root (t - 2), over regular (t - 3)

g(x)= square root (x - 4) plus square root (r) minus 2

I don't know how to use latex, so this is the best I can do. Sorry if it's unclear.
I don't understand how to find the domain.

2. It's interesting that you took the time to find out that mathematical typesetting is done using LaTeX but you didn't bother to at least learn some...

Anyway

$g(t) = \frac{\sqrt{t-2}}{t-3}$.

You can't have the square root of a negative number, so $t - 2 \geq 0$.

Therefore $t \geq 2$.

Also you can't have a denominator of $0$, so that means

$t - 3 \neq 0$

$t \neq 3$.

Putting it together gives

$t \in [2, 3) \cup (3, \infty)$.

3. is your second question this??
$g(x) = \sqrt{x-4} + \sqrt{x} - 2 ??$

if this was your second question then the answer is $x \geq 4$ and $x \geq 0$ therefore the final answer is $x \in [4,\infty)$

if your question is $g(x) = \sqrt{x-4} + \sqrt{r} - 2 ??$

then the answer is $x \geq 4$ and $r \geq 0$

the domain for x is $x \in [4,\infty)$
and the domain for r is $r \in [0,\infty)$

4. Originally Posted by grgrsanjay
is your second question this??
$g(x) = \sqrt{x-4} + \sqrt{x} - 2 ??$

if this was your second question then the answer is $x \geq 4$ and $x \geq 0$ therefore the final answer is $x \in [0,\infty)$
Actually it would be $x \in [4, \infty)$.

5. ah just was a silly mistake