# How do you find the domain and range of g(x)=sqrt((x+3)/(x-2))?

Aug 19, 2017

Range: $\text{ } x \in \left(- \infty , - 3\right] \cup \left(2 , + \infty\right)$

Domain: $\text{ } g \left(x\right) \in \left[0 , 1\right) \cup \left(1 , + \infty\right)$

#### Explanation:

In order to find the domain of this function, you need to find all the values that $x$ can take for which $g \left(x\right)$ is defined.

Right from the start, you know that the denominator of the fraction cannot be equal to $2$ because that would make the function undefined.

$x - 2 \ne 0 \implies x \ne 2$

Now, you know that when working with real numbers, you can only take the square root of a positive number.

This implies that you must have

$\frac{x + 3}{x - 2} \ge 0$

You know that when $x = - 3$, you have

$\frac{- 3 + 3}{- 3 - 2} = \frac{0}{- 5} \ge 0$

so you can say that $x = \left\{- 3\right\}$ will be included in the domain of the function.

Now, in order to have

$\frac{x + 3}{x - 2} > 0$

you must look at two possible situations

$\textcolor{w h i t e}{a}$

• $x + 3 > 0 \text{ " ul(and) " } x - 2 > 0$

In this case, you must have

$x + 3 > 0 \implies x \ge - 3$

and

$x - 2 > 0 \implies x > 2$

This implies that the solution interval will be

$\left(- 3 , + \infty\right) \cap \left(2 , + \infty\right) = \left(2 , + \infty\right)$

This tells you that any value of $x$ that is greater than $2$ will satisfy the inequality $\frac{x + 3}{x - 2} > 0$.

$\textcolor{w h i t e}{a}$

• $x + 3 < 0 \text{ "ul(and)" } x - 2 < 0$

In this case, you must have

$x + 3 < 0 \implies x < - 3$

and

$x - 2 < 0 \implies x < 2$

This implies that the solution interval will be

$\left(- \infty , - 3\right) \cap \left(- \infty , 2\right) = \left(- \infty , - 3\right)$

This tells you that any value of $x$ that is less than $- 3$ will also satisfy the inequality $\frac{x + 3}{x - 2} > 0$.

Therefore, the domain of the function will be--remember that $x = - 3$ is also included in the domain!

$\text{domain: } \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{x \in \left(- \infty , - 3\right] \cup \left(2 , \infty\right)}}}$

This tells you that any value of $x$ that is less than or equal to $- 3$ or greater than $2$ will get you

$\frac{x + 3}{x - 2} \ge 0$

Now, to find the range of the function, you must determine the values that $g \left(x\right)$ can take for any value of $x$ that is part of its domain.

Since you're working with real numbers, you can say that taking the square root of a positive number will always return a positive number.

$g \left(x\right) \ge 0$

You know that when $x = - 3$, you have

$g \left(- 3\right) = \sqrt{\frac{- 3 + 3}{- 3 - 2}} = 0$

Now, it's important to realize that the range of the function will not include $1$ because you can never have

$\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} + 3 = \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} - 2$

$3 \ne - 2$

This means that you don't have a value of $x$ for which $\frac{x + 3}{x - 2} = 1$.

Therefore, the range of the function will be

$\text{range: } \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{g \left(x\right) \in \left[0 , 1\right) \cup \left(1 , + \infty\right)}}}$

graph{sqrt( (x+3)/(x-2)) [-16.02, 16.01, -8.01, 8]}