# Thread: Finding the domain of a function

1. ## Finding the domain of a function

$f(x)=x/\sqrt{39x^2-7}$

For clarification purposes, in word form this would be: the fraction of x over the square root of 39 x squared minus seven.

It would help alot if someone could help me find and write the answer Set Builder Notation (ex: [-5, infinity)). Thanks in advance for your assistance.

2. Originally Posted by currypuff
$f(x)=x/\sqrt{39x^2-7}$

For clarification purposes, in word form this would be: the fraction of x over the square root of 39 x squared minus seven.

It would help alot if someone could help me find and write the answer Set Builder Notation (ex: [-5, infinity)). Thanks in advance for your assistance.
The domain is all values of $x$ that the function takes on. In this scenario, not only must the square root not be negative, but it cannot be zero, so:

$39x^2-7>0$

$39x^2>7$

$x^2>\frac{7}{39}$

$x>\sqrt{\frac{7}{39}}$
$x<-\sqrt{\frac{7}{39}}$

$x: \left(-\infty, -\sqrt{\frac{7}{39}}\right) \cup \left(\sqrt{\frac{7}{39}}, \infty \right)$

3. Originally Posted by colby2152
The domain is all values of $x$ that the function takes on. In this scenario, not only must the square root not be negative, but it cannot be zero, so:

$39x^2-7>0$

$39x^2>7$

$x^2>\frac{7}{39}$

$x>\sqrt{\frac{7}{39}}$
$x<-\sqrt{\frac{7}{39}}$

$x: (-\infty, -\sqrt{\frac{7}{39}}) \cap (\sqrt{\frac{7}{39}}, \infty)$
use \left( and \right) so that the brackets encompass everything. so you would get:

$x: \left(-\infty, -\sqrt{\frac{7}{39}} \right) \cap \left(\sqrt{\frac{7}{39}}, \infty \right)$

which looks nicer

...well, sort of, that answer's ugly either way...

4. Originally Posted by Jhevon
use \left( and \right) so that the brackets encompass everything. so you would get:

$x: \left(-\infty, -\sqrt{\frac{7}{39}} \right) \cap \left(\sqrt{\frac{7}{39}}, \infty \right)$

which looks nicer

...well, sort of, that answer's ugly either way...
it should be $x: \left(-\infty, -\sqrt{\frac{7}{39}} \right) \cup \left(\sqrt{\frac{7}{39}}, \infty \right)$ ?

5. Originally Posted by curvature
it should be $x: \left(-\infty, -\sqrt{\frac{7}{39}} \right) \cup \left(\sqrt{\frac{7}{39}}, \infty \right)$ ?
yes, that's right. the union.

thanks for spotting it, i didn't see that, i just copied the code Colby used

6. Originally Posted by Jhevon
yes, that's right. the union.

thanks for spotting it, i didn't see that, i just copied the code Colby used
Cup and cap!

How does it look now Jhevon?

7. Originally Posted by colby2152
Cup and cap!

How does it look now Jhevon?
looks good!

(though i would use $\in$ instead of : but the answer is understood, so no big deal)