Since this function involves a square root (and the number inside the square root, x^2-8 in this case, can never be negative in the real number plane), this means that the lowest possible value that x^2-8 can be is 0.
x^2-8 can never be negative because two real numbers cannot ever be squared to make a negative number, only ever a positive number or 0.
Therefore, since you know that the value of x^2-8 must be greater than or equal to 0, you can set up the equation x^2-8≥0.
Solve for x and you will get sqrt(8), or 2sqrt(2) when simplified, as the domain (all possible real values of x). Therefore, x≥2sqrt(2) (or [2sqrt(2),oo).
For the range, since you know that x^2-8≥0, then sqrt(x^2-8) must be ≥0. If you substitute x^2-8 with 0, then you will get the range of y≥0 or [0,oo).
Hope this helps!
