# How do you find domain and range for  f(x) =sqrt(4-3x) + 2?

Jul 5, 2018

Domain: $\left(- \infty , \frac{4}{3}\right]$

Range: $\left[2 , \infty\right)$

#### Explanation:

We know whatever we have under the radical cannot be negative, so we can set the following inequality:

$4 - 3 x \ge 0$

Adding $3 x$ to both sides gives us

$3 x \le 4$

$x \le \frac{4}{3}$ as our domain or $\left(- \infty , \frac{4}{3}\right]$ in interval notation.

The lowest value that this expression can take on is $2$, when the radical just evaluates to zero. Our range can be $2$ or greater.
$\left[2 , \infty\right)$ in interval notation