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

Jul 7, 2018

$x \ge 3$ or

in interval notation $\left[3 , \infty\right)$

#### Explanation:

Given: $F \left(x\right) = \sqrt{x - 3}$

A function starts out having a domain of all Reals $\left(- \infty , \infty\right)$

A square root limits the function because you can't have negative numbers under the square root (they are called imaginary numbers).

This means $\text{ } x - 3 \ge 0$

Simplifying: $\text{ } x \ge 3$

Jul 7, 2018

The domain is $x \in \left[3 , + \infty\right)$. The range is $y \in \left[0 , + \infty\right)$

#### Explanation:

Let $y = \sqrt{x - 3}$

What's under the sqrt sign must be $\ge 0$

Therefore,

$x - 3 \ge 0$

$\implies$, $x \ge 3$

The domain is $x \in \left[3 , + \infty\right)$

When $x = 3$, $y = \sqrt{3 - 3} = 0$

And

${\lim}_{x \to + \infty} y = {\lim}_{x \to + \infty} \sqrt{x - 3} = + \infty$

Therefore,

The range is $y \in \left[0 , + \infty\right)$

graph{sqrt(x-3) [-12.77, 27.77, -9.9, 10.38]}