# Domain of this funtion is requested:   Not sure if it's pasted! so: it is a fraction which on the nominator, there is logarithm of the root which under the root is: x^2-3 and on the...

Domain of this funtion is requested:

Not sure if it's pasted! so: it is a fraction which on the nominator, there is logarithm of the root which under the root is: x^2-3

and on the denominator, there is an absolute value: | x^2-4 |

for finding the domain of this function I solved these conditions:

for the nominator:

1) logarithm of the root which under the root is: x^2-3 >= 0 which the values of x's, solved as: x =< -2 V x >= 2

2) the root which under the root is: x^2-3 > 0 which the values of x's, solved as: x < -(root)3 V x > (root)3

3) x^2-3 >= 0 which the values of x's, solved as: x =< -(root)3 V x >= (root)3

and, for the denominator:

1) | x^2-4 | "is not equal to" 0 so x "is not equal to" (-/+) 2

2) | x^2-4 | > 0 which the values of x's, solved as: x < -2 V x > 2 V -2 < x < 2

at the end with getting the intersection of all the five conditions, I got the domain as: (-infinite, -2) U (2, +infinite) which the correct answer is (-infinite, -2) U (-2, -(root)3) U (root 3, 2) U (2, +infinite)

I tried to verify what was my mistake, and I only could see if I've removed the first condition of the nominator, I would get the same answer, but if the method for getting the values of a fraction is to solve the nominator and the denominator separately when they are considering to be greater and equal to zero, then I couldn't find the reasoning to remove the first condition of the nominator! I want to fully understand how I supposed to solve domains!

Posted on

I have checked how you've solved the problem. There is one mistake.

When you have tried to solve the inequality `x^2-3 gt= 0` , you have get wrong answers.

Let's see what are the values that cancel out the equation and where your mistake has happened.

`x^2 - 3 = 0.`

`x^2 = 3 =gt x_1 = +sqrt 3 and x_2 = -sqrt 3`

For no reason, you have put instead of the values +- sqrt 3, the values +- 2. This was the point where the mistake has happened.

Choose a value for x, which is found between the roots.

`x = 0 =gtx^2 - 3 = 0 - 3 = -3`

Therefore, x^2 - 3 is positive for values that are found outside -sqrt3 and sqrt3.

`x^2 - 3 gt= 0 <=> x in (-oo ; -sqrt3)U(sqrt3 ; +oo)`

Considering the conditions that come with the denominator, the final answer is `x in (-oo, -2)U(-2 , -sqrt 3)U(sqrt 3 , 2)U(2 , +oo).`