Finding the domain of a function

**currypuff** · January 29th 2008, 07:08 AM

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

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.

**colby2152** · January 29th 2008, 07:26 AM

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:

$39x^2-7>0$

$39x^2>7$

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

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

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

**Jhevon** · January 29th 2008, 04:16 PM

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...

**curvature** · January 29th 2008, 08:51 PM

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

**Jhevon** · January 29th 2008, 09:15 PM

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

**colby2152** · January 30th 2008, 04:42 AM

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?

**Jhevon** · January 30th 2008, 03:51 PM

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)

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