# What is the domain and range of y =sqrt(3x - 5)?

Dec 17, 2017

Domain: $\left\{x | \frac{5}{3} \le x\right\}$
Range: $y \in {\mathbb{R}}_{\ge 0}$

#### Explanation:

To find the domain, we just need to look at where the function is defined with real numbers. The only thing that can make this function not defined is if the value inside the square root were to be negative. To find out when this is the case, we need to solve the following inequality:

$3 x - 5 < 0$

$3 x - \cancel{5 + 5} < 0 + 5$

$\frac{\cancel{3} x}{\cancel{3}} < \frac{5}{3}$

$x < \frac{5}{3}$

So, we can conclude that the function is not defined when $x$ is less than $5 \frac{\setminus}{3}$. This means that the function is defined when $x$ is greater than or equal to $5 \frac{\setminus}{3}$:
$\left\{x | \frac{5}{3} \le x\right\}$

The range goes from $0$ to $\infty$ (mind you, square roots can never have negative answers), and the function is continuous on that interval, so the range must be all the positive real numbers and $0$, ${\mathbb{R}}_{\ge 0}$