Question

Domain of f(x)?

`f(x)=sqrt(x^2-2)/sqrt(x^2-9)`

Answer 1

graph{sqrt(x^2-2)/sqrt(x^2-9) [-10, 10, -5, 5]}

The domain in interval notation is - `(-oo, -3)uu(3,oo)`

Explanation:

Solve this by checking which values make the function undefined. Under the square root, nothing can be negative. In the denominator, nothing can be zero. If x=3, the denominator becomes 0, so x must be bigger than 3. However anything smaller than -3 also works, since x is squared. So, the function is undefined between -3 and 3, and defined everywhere else, making the domain `(-oo, -3)uu(3,oo)` in interval notation. It can also be written like this: `x<-3 or x>3`

Note that in both 3 and -3 are not included, since the function is defined at both those values.

Answer 2

The domain must prevent the arguments of the square roots from becoming negative:

`x^2-2>=0` and `x^2-9>=0`

`x^2>=2` and `x^2>=9`

Also the denominator must not become 0, therefore, the second equation must not allow equality:

`x^2>=2` and `x^2>9`

Logically, the above becomes `x^2>9`

Because of the way that the square root works, we obtain two regions:

`x < -3` or `x > 3`

The above is the domain.