 # Domain of a Square Root Function

How to find the domain of square root functions? Several examples are presented along with detailed solutions and explanations with graphical explanations and interpretation of the domain.

## How to find the domain of a square root function?

We first need to understand that $\sqrt x$ is real only if the argument $x$ which is the quantity under the radical $\sqrt{}$ satisfies the condition: $x \ge 0$. This can easily be verified by examining the graph of $y = \sqrt x$ shown below: The graph exists only for $x \ge 0$. ### Example 1

Find the domain of the function $f(x) = \sqrt{x - 2}$.

### Solution:

$f(x) = \sqrt{x - 2}$ takes real values if the argument $x - 2$, which is the quantity under the radical $\sqrt{}$, satifies the condition: $x - 2 \ge 0$. The solution of the inequality is
$x \ge 2$
which is the domain of the function and this can be checked graphically as shown below where the graph of $f$ "exists" for $x \ge 2$. ### Example 2

Find the domain of the function $f(x) = \sqrt{|x - 1|}$.

### Solution

$f(x) = \sqrt{|x - 1|}$ takes real values if the argument $|x - 1|$, which is the quantity under the radical $\sqrt{}$, satifies the condition: $|x - 1| \ge 0$. The solution of the inequality is
all real number because the absolute value expression $|x - 1|$ is always positive or zero for $x = 1$.
The domain of the function is the set of real numbers $\mathbb{R}$ and this can be checked graphically as shown below where the graph of $f$ "exists" for all $x$ values.. ### Example 3

Find the domain of the function $f(x) = \dfrac{1}{ \sqrt{x + 3}}$.

### Solution

Taking into account that the function is the ratio of two functions and that division by zero is not allowed, the given function takes real values if the argument $x + 3$, which is the quantity under the radical $\sqrt{}$, satisfies the condition: $x + 3 \gt 0$. Note the symbol of the inequality used is $\gt$ and not $\ge$ because we do not want to have zero in the denominator. Solution of the inequality is
$x \gt - 3$.
The domain of the function is the set of real numbers greater than -3 and this can be checked graphically as shown below where the graph of $f$ "exists" for all $x > - 3$ values.. ### Example 4

Find the domain of the function $f(x) = \dfrac{\sqrt{x + 4}}{ \sqrt{x - 2}}$.

### Solution

The given function takes real values if two conditions are satisfied.
1) $x + 4 \ge 0$ , zero is allowed in the numerator, hence the use of the inequality symbol $\ge$.
and
2) $x - 2 \gt 0$ , zero is not allowed in the denominator, hence the use of the inequality symbol $\gt$.
The domain of the function is the intersection of the two solution sets of the two inequalities above..
$x \ge - 4$ and $x \gt 2$
The intersection of the two solution sets above is..
$x \gt 2$
Which is the domain of the given function as shown below in the graph of $f$.. ### Example 5

Find the domain of the function $f(x) = \sqrt{\dfrac{x + 4}{ {x - 2}}}$.

### Solution

The given function takes real values if
$\dfrac{x + 4}{ {x - 2}} \ge 0$
We need to solve the above inequality. The zeros of the numerator and denominator are
x = - 4 and x = 2
The zeros split the number line into 3 intervals over which the sign of the inequality is the same. Hence the intervals.
$(-\infty , -4)$ , $(-4 , 2 )$ , $(2 , \infty)$
We select a value within each interval and use it to find the sign of the expression $\dfrac{x + 4}{ {x - 2}}$.
1) On the interval $(-\infty , -4)$ , select x = -6 and substitute x in $\dfrac{x + 4}{ {x - 2}}$ by -6 to determine its sign
$\dfrac{ -6 + 4}{ {-6 - 2}} \gt 0$
2) On the interval $(-4 , 2)$ , select x = 0 and substitute x in $\dfrac{x + 4}{ {x - 2}}$ by 0 to determine its sign
$\dfrac{ 0 + 4}{ {0 - 2}} \lt 0$
3) On the interval $(2 , \infty)$ , select x = 3 and substitute x in $\dfrac{x + 4}{ {x - 2}}$ by 3 to determine its sign
$\dfrac{ 3 + 4}{ {3 - 2}} \gt 0$
Hence the domain is the union of all intervals over which $\dfrac{x + 4}{ {x - 2}} \gt 0$ and is given by.
$(-\infty , -4] \cup (2 , \infty)$
The graph of $f$ is as shown below and we may easily check the domain found above. Note x = 2 is not the domain because the division by zero is not allowed.. ### Example 6

Find the domain of the function $f(x) = \sqrt{-x^2-4}$.

### Solution

The given function takes real values if
$-x^2 - 4 \ge 0$
The expression $x^2 + 4$ is the sum of a square and a positive number. Hence
$x^2 + 4 \ge 0$
Multiply all terms of the above inequality by -1 and change the symbol of inequality to obtain
$- x^2 - 4 \le 0$
Hence the domain of the given function is an empty set and the given function has no graph. Try to graph it using a graphing calculator.

### Example 7

Find the domain of the function $f(x) = \dfrac{\sqrt{6 - x}}{ \sqrt{x - 2}}$.

### Solution

The given function takes real values if two conditions are satisfied.
1) $6 - x \ge 0$ , zero is allowed in the numerator, hence the use of the inequality symbol $\ge$.
and
2) $x - 2 \gt 0$ , zero is not allowed in the denominator, hence the use of the inequality symbol $\gt$.
The domain of the function is the intersection of the two solution sets of the two inequalities above..
$x \le 6$ and $x \gt 2$
The intersection of the two solution sets above is given by the interval.
$(2 , 6]$
The graph of $f$ is as shown below and we may easily check the domain found above. Note x = 2 is not the domain because the division by zero is not allowed.. ### Example 8

Find the domain of the function $f(x) = \sqrt{x^2 - 4}$.

### Solution

The given function takes real values if
$x^2 - 4 \ge 0$ which can also be written $(x - 2)(x + 2) \ge 0$
The solution set to the above quadratic inequality is given by the interval
$(-\infty , -2] \cup [2 , \infty)$
The graph of $f$ is as shown below and we may easily check the domain found above.. ### Example 9

Find the domain of the function $f(x) = \sqrt{4 - x^2}$.

### Solution

The given function takes real values if
$4 - x^2 \ge 0$ which can also be written as $(2 - x)(2 + x) \ge 0$
The solution set to the above quadratic inequality is given by the interval
$[2 , 2]$
The graph of $f$ is as shown below and we may easily check the domain found above.. 