Finding the domain

Printable View

February 10th 2010, 07:28 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.
February 10th 2010, 07:35 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 $R=(-\infty,\infty)$
February 10th 2010, 07:37 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
February 10th 2010, 07:45 PM
Pulock2009

x>sqrt(3)/2-1/2 should be the answer.
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.
February 10th 2010, 07:50 PM
General

Quote:

Originally Posted by Pulock2009

x>sqrt(3)/2-1/2 should be the answer.
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.

You have a sign mistake.
It should be ${\color{red}-}\frac{3}{4}$ .
February 10th 2010, 09:48 PM
Mathlv

finding the domain

Thank you for the help General, danielomalmsteen, and pulock2009. I cannot believe that I have forgotten about applying the "completing the square" to get to the domain.