Domain of f(x)?

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

2 Answers
Jan 12, 2018

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.

Jan 12, 2018

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.