A step-by-step tutorial, with detailed solutions, on how to find the domain of square root functions is presented. You can also use our Step-by-Step Calculator to Find Domain of a Function.
Definition of the Domain of a Function: For a function \( f \) defined by an expression with variable \( x \), the implied domain of \( f \) is the set of all real numbers variable \( x \) can take such that the expression defining the function is real. The domain can also be given explicitly.
For a square root function given by \( f(x) = \sqrt{x} \) to have real values, the radicand \( x \) must be greater than or equal to zero.
Examples on How to Find the Domain of Square Root Functions
Example 1
Find the domain of function \( f \) defined by:
\[ f(x) = \sqrt{x - 1} \]View Solution
For \( f(x) \) to have real values, the radicand (expression under the radical) of the square root function must be positive or equal to 0. Hence:
\[ x - 1 \geq 0 \]The solution set to the above inequality is the domain of \( f(x) \) and is given by \( x \geq 1 \), or in interval form:
\[ [1, +\infty) \]Example 2
Find the domain of function \( f \) defined by:
\[ f(x) = \sqrt{(x - 2)(x + 3)} \]View Solution
For \( f(x) \) to have real values, the radicand \((x - 2)(x + 3)\) must be greater than or equal to zero. Hence:
\[ (x - 2)(x + 3) \geq 0 \]Solve the above inequality to obtain the solution set, which is also the domain, in interval form as follows:
\[ (-\infty, -3] \cup [2, +\infty) \]Example 3
Find the domain of function \( f \) defined by:
\[ f(x) = \sqrt{x^2 + 2x - 1} \]View Solution
For \( \sqrt{x^2 + 2x - 1} \) to be real, the radicand must be positive or equal to 0. Hence the inequality:
\[ x^2 + 2x - 1 \geq 0 \]The solution set of the above quadratic inequality, which is also the domain, is given in interval form as follows:
\[ (-\infty, -1 - \sqrt{2}] \cup [-1 + \sqrt{2}, +\infty) \]Example 4
Find the domain of function \( f \) defined by:
\[ f(x) = \sqrt{\dfrac{2x - 1}{x + 3}} \]View Solution
The domain of this function is the set of all values of \( x \) such that:
\[ \dfrac{2x - 1}{x + 3} \geq 0 \]Solving this inequality gives the solution set, which is also the domain:
\[ (-\infty, -3) \cup \left[\frac{1}{2}, +\infty\right) \]Example 5
Find the domain of function \( f \) defined by:
\[ f(x) = \sqrt{|-2x - 6|} \]View Solution
The domain of this function is the set of all values of \( x \) such that \( |-2x - 6| \geq 0 \). We need to solve the inequality:
\[ |-2x - 6| \geq 0 \]Because of the absolute value, the expression \( |-2x - 6| \) is greater than or equal to 0 for all real numbers. Hence, the domain of the above function is given by:
\[ (-\infty, +\infty) \]