# Finding the domain

• Feb 10th 2010, 08:28 PM
Mathlv
Finding the domain
Hello,
I was asked to find the domain of this function
f(x)= $\frac{1}{{\sqrt {x^2 + x + 1} }}
$
. So far I cannot factor out the denominator. I know that the denominator cannot be = 0,
but I am not able to find the x value where

$
\sqrt {x^2 + x + 1} > 0
$

Can someone tell me if there is a domain or domains in this function? Thank you for the help.
• Feb 10th 2010, 08:35 PM
General
Quote:

Originally Posted by Mathlv
Hello,
I was asked to find the domain of this function
f(x)= $\frac{1}{{\sqrt {x^2 + x + 1} }}
$
. So far I cannot factor out the denominator. I know that the denominator cannot be = 0,
but I am not able to find the x value where

$
\sqrt {x^2 + x + 1} > 0
$

Can someone tell me if there is a domain or domains in this function? Thank you for the help.

You have a mistake.
Inside the root must be bigger than the zero.
not the root itself.
Actually, It will give the same result, but the correct thing is taking inside root bigger than zero.
$x^2+x+1 > 0$
Complete the square, to get:

$(x+\frac{1}{2})^2+\frac{3}{4}>0$

$(x+\frac{1}{2})^2>-\frac{3}{4}$

Which is true for all values of x
since any squared things is bigger than or equal to zero.
Hence: The domian is $R=(-\infty,\infty)$
• Feb 10th 2010, 08:37 PM
danielomalmsteen
Another way is to go to discriminant of the quadratic equation. Clearly this is negative, then the associated parabola takes only positive values and it follows that the domain is all real
• Feb 10th 2010, 08:45 PM
Pulock2009
make a square by introducing a (1/2)^2 into the expression. then u should get the following intermediate step:(x+1/2)^2>3/4.
• Feb 10th 2010, 08:50 PM
General
Quote:

Originally Posted by Pulock2009
It should be ${\color{red}-}\frac{3}{4}$.