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

Feb 1, 2018

${D}_{f} = \left\{x \le 8\right\}$ and

R_f = { 0<=y<∞}  where $y = {\left(8 - x\right)}^{\frac{1}{2}}$

#### Explanation:

Since the given function is a square root function so the only NON-NEGATIVE values are allowed in the Domain and the Range will also be having NON-NEGATIVE Real numbers only.

For Domain, ${D}_{f}$ :

$8 - x \ge 0$
$\Rightarrow x - 8 \le 0$
$\Rightarrow x \le 8$

and we can see that this function will acquire every positive value and Zero also (when $x = 8$).

Thus the Range of function, R_f = { 0<=y<∞}