How do you find the domain of f(x)=sqrt((x^2-16))?

1 Answer
Apr 26, 2017

x >=4 and x<=-4

Explanation:

Domain for a function means the value/s of x for which the function is valid. In this case, the function is invalid if (x^2-16) is < 0 as there are no real values of the function. So, you find the values of x which would make (x^2-16) greater or equal to 0 as the function would be valid.
So,

(x^2-16) >= 0
x^2 >= 16
x^2 >= 4^2
|x| >= 4
Either,

x >= 4

Or,

x <= -4

So for the values of x >=4 and x<=-4 , (x^2-16) will be greater or equal to zero which will be give some value under square root making the function valid.
The remaining values -4 < x < 4, would give you a negative term for (x^2-16) which would not be valid under square root making the function invalid.

SO the domain is x >=4 and x<=-4