# 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.

## 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 $$x \ge 0$$.

## Examples with their Solutions

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.