# How do you find the domain and range for f(x)=sqrt( 3-8x)?

May 17, 2017

The domain is $x \in \left(- \infty , \frac{3}{8}\right]$
The range is $f \left(x\right) \in \left[0 , + \infty\right)$

#### Explanation:

$f \left(x\right) = \sqrt{3 - 8 x}$

What's under the square root sign is $\ge 0$

Therefore,

$3 - 8 x \ge 0$

$8 x \le 3$

$x \le \frac{3}{8}$

so,

The domain of $f \left(x\right)$ is $x \in \left(- \infty , \frac{3}{8}\right]$

when $x = \frac{3}{8}$ , $f \left(\frac{3}{8}\right) = 0$

and when $x = - \infty$, $f \left(- \infty\right) = + \infty$

So the range is $f \left(x\right) \in \left[0 , + \infty\right)$