How do you find the domain and range of #f(x)= sqrt(x² - 8) #?

1 Answer
Apr 12, 2018

Answer:

The domain is #x≥2sqrt(2)# (or #[2sqrt(2),oo)# and the range is #y≥0# or #[0,oo)# .

Explanation:

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!