Finding the Domain (really simple)

• Sep 3rd 2010, 09:24 PM
Mariolee
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. :p
I don't understand how to find the domain.
• Sep 3rd 2010, 09:58 PM
Prove It
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

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

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

Therefore $\displaystyle t \geq 2$.

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

$\displaystyle t - 3 \neq 0$

$\displaystyle t \neq 3$.

Putting it together gives

$\displaystyle t \in [2, 3) \cup (3, \infty)$.
• Sep 4th 2010, 04:23 AM
grgrsanjay
$\displaystyle g(x) = \sqrt{x-4} + \sqrt{x} - 2 ??$

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

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

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

the domain for x is $\displaystyle x \in [4,\infty)$
and the domain for r is $\displaystyle r \in [0,\infty)$
• Sep 4th 2010, 06:27 AM
Prove It
Quote:

Originally Posted by grgrsanjay
$\displaystyle g(x) = \sqrt{x-4} + \sqrt{x} - 2 ??$
if this was your second question then the answer is $\displaystyle x \geq 4$ and $\displaystyle x \geq 0$ therefore the final answer is $\displaystyle x \in [0,\infty)$
Actually it would be $\displaystyle x \in [4, \infty)$.