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

Jul 2, 2018

Domain: $\left[3 , \infty\right) \text{ or } x \ge 3$

Range: $\left[- \sqrt{6} , 0\right) \text{ or } - \sqrt{6} \le y < 0$

#### Explanation:

Given: $y = \sqrt{x - 3} - \sqrt{x + 3}$

Both the domain is the valid inputs $x$. The range is the valid outputs $y$.

Since we have two square roots, the domain and the range will be limited.

$\textcolor{b l u e}{\text{Find the Domain:}}$

The terms under each radical must be $\ge 0$:

x - 3 >= 0; " " x + 3 >= 0

x >= 3; " "x >= -3

Since the first expression must be $\ge 3$, this is what limits the domain.

Domain: $\left[3 , \infty\right) \text{ or } x \ge 3$

$\textcolor{red}{\text{Find the Range:}}$

The range is based on the limited domain.

Let $x = 3 \implies y = \sqrt{3 - 3} - \sqrt{3 + 3} = - \sqrt{6}$

Let $x = 100 \implies y = \sqrt{97} - \sqrt{103} \approx - .3$

Let $x = 1000 \implies y = \sqrt{997} - \sqrt{1003} \approx - .09$

$x \to \infty , y \to 0$

Range: $\left[- \sqrt{6} , 0\right) \text{ or } - \sqrt{6} \le y < 0$

Jul 2, 2018

The domain is $x \in \left[3 , + \infty\right)$. The range is $y \in \left[- \sqrt{6} , {0}^{-}\right)$

#### Explanation:

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

$\implies$, $x - 3 \ge 0$ and $x + 3 \ge 0$

$\implies$, $\left\{\begin{matrix}x \ge 3 \\ x \ge - 3\end{matrix}\right.$

Therefore,

The domain is $\left(x \ge 3\right) \cap \left(x \ge - 3\right)$

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

When $x = 3$, $\implies$, $y = 0 - \sqrt{6}$

And when $x \to + \infty$

${\lim}_{x \to + \infty} y = {0}^{-}$

Therefore,

The range is $y \in \left[- \sqrt{6} , {0}^{-}\right)$

graph{sqrt(x-3)-sqrt(x+3) [-1.42, 18.58, -6.36, 3.64]}

Jul 2, 2018

Domain: $\left[3 , \infty\right)$

Range: $\left[- \sqrt{6} , 0\right)$

#### Explanation:

Given:

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

First note that the square roots are well defined and real if and only if $x - 3 \ge 0$ and $x + 3 \ge 0$. Hence it is necessary and sufficient that $x \ge 3$.

So the domain of the function is $\left[3 , \infty\right)$

To find the range, note that when $x = 3$ then:

$y = \sqrt{\left(\textcolor{b l u e}{3}\right) - 3} - \sqrt{\left(\textcolor{b l u e}{3}\right) + 3} = \sqrt{0} - \sqrt{6} = - \sqrt{6}$

We find:

${\lim}_{x \to \infty} \left(\sqrt{x - 3} - \sqrt{x + 3}\right) = {\lim}_{x \to \infty} \frac{\left(\sqrt{x - 3} - \sqrt{x + 3}\right) \left(\sqrt{x - 3} + \sqrt{x + 3}\right)}{\sqrt{x - 3} + \sqrt{x + 3}}$

$\textcolor{w h i t e}{{\lim}_{x \to \infty} \left(\sqrt{x - 3} - \sqrt{x + 3}\right)} = {\lim}_{x \to \infty} \frac{\left(x - 3\right) - \left(x + 3\right)}{\sqrt{x - 3} + \sqrt{x + 3}}$

$\textcolor{w h i t e}{{\lim}_{x \to \infty} \left(\sqrt{x - 3} - \sqrt{x + 3}\right)} = {\lim}_{x \to \infty} \frac{- 6}{\sqrt{x - 3} + \sqrt{x + 3}}$

$\textcolor{w h i t e}{{\lim}_{x \to \infty} \left(\sqrt{x - 3} - \sqrt{x + 3}\right)} = 0$

Note that $- \frac{6}{\sqrt{x - 3} + \sqrt{x + 3}}$ is continuous and monotonically increasing.

Hence the range of the given function runs from the minimum value $- \sqrt{6}$ up to but not including the limit $0$.

That is, the range is $\left[- \sqrt{6} , 0\right)$

graph{y = sqrt(x-3)-sqrt(x+3) [-10, 10, -5, 5]}