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

Apr 12, 2018

The domain is x≥2sqrt(2) (or $\left[2 \sqrt{2} , \infty\right)$ and the range is y≥0 or $\left[0 , \infty\right)$ .

#### 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 $2 \sqrt{2}$ when simplified, as the domain (all possible real values of x). Therefore, x≥2sqrt(2) (or $\left[2 \sqrt{2} , \infty\right)$.

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 $\left[0 , \infty\right)$.

Hope this helps!