# What is the domain and range for f(x)=sqrt(2x-8)?

Sep 21, 2015

The domain is: $\left[4 , \infty\right)$
The range is: $\left(0 , \infty\right)$

#### Explanation:

$f \left(x\right) = \sqrt{2 x - 8} \implies$ square root of negative number is not real:
hence the domain is calculated as follow:
$2 x - 8 \ge 0$
$2 x \ge 8$
$x \ge 4$
In interval form:
$\left[4 , \infty\right)$
Now to find the range, consider the possible minimum and the maximum values of y for the valid domain, so in this case the range is all positive real numbers:
$y > 0$
$\left(0 , \infty\right)$