A step-by-step tutorial, with detailed solutions, on how to find the domain of rational 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 that variable \( x \) can take such that the expression defining the function is real. The domain can also be given explicitly.
Find the Domain of a Rational Function: Examples with Solutions
Example 1
Find the domain of the function \( f \) defined by:
\[ f(x) = \dfrac{1}{x - 2} \]View Solution
\( f(x) \) can take real values if the denominator of \( f(x) \) is NOT ZERO because division by zero is not allowed in mathematics:
\[ x - 2 \neq 0 \]Solve the inequality for \( x \) to obtain the domain restriction: \( x \neq 2 \).
Which in interval form may be written as follows:
\[ (-\infty, 2) \cup (2, +\infty) \]Example 2
Find the domain of the function \( f \) defined by:
\[ f(x) = \dfrac{x + 3}{x^2 + 7} \]View Solution
For \( f(x) \) to have real values, the denominator must be different from zero. Hence:
\[ x^2 + 7 \neq 0 \]The expression \( x^2 + 7 \) is always positive (a square added to a positive number). Hence, the domain of \( f \) is given by the interval:
\[ (-\infty, +\infty) \]Example 3
Find the domain of the function \( f \) defined by:
\[ f(x) = \dfrac{2x + 9}{2x^2 + x - 15} \]View Solution
For \( f(x) \) given above to be real, its denominator must be different from zero. Let us first find the roots of the denominator by solving the equation:
\[ 2x^2 + x - 15 = 0 \]The roots are \( -3 \) and \( \dfrac{5}{2} \).
The denominator \( 2x^2 + x - 15 \) is not equal to zero for all real values except \( -3 \) and \( \dfrac{5}{2} \). Hence, the domain of the given function is given by:
\[ (-\infty, -3) \cup \left(-3, \frac{5}{2}\right) \cup \left(\frac{5}{2}, +\infty\right) \]Example 4
Find the domain of the function \( f \) given by:
\[ f(x) = \dfrac{2}{2x - 6} - \dfrac{x}{4x + 7} \]View Solution
For \( f(x) \) to be real, both denominators \( 2x - 6 \) and \( 4x + 7 \) must not be equal to zero. Let us find the values of \( x \) that make the two denominators equal to zero:
\( 2x - 6 = 0 \) gives \( x = 3 \)
\( 4x + 7 = 0 \) gives \( x = -\dfrac{7}{4} \)
\( f(x) \) is real for all real values except \( 3 \) and \( -\dfrac{7}{4} \). The domain of the above function is given by:
\[ \left(-\infty, -\frac{7}{4}\right) \cup \left(-\frac{7}{4}, 3\right) \cup (3, +\infty) \]Example 5
Find the domain of the function \( f \) defined by:
\[ f(x) = \dfrac{-5x + 6}{x^3 - x} \]View Solution
Let us first find the values that make the denominator equal to zero:
\[ x^3 - x = 0 \]Factor the expression on the left-hand side of the equation:
\[ x(x^2 - 1) = 0 \implies x(x + 1)(x - 1) = 0 \]Solve the above equation to obtain the solution set: \( x = 0 \), \( x = 1 \), and \( x = -1 \).
The domain is given in interval form as follows:
\[ (-\infty, -1) \cup (-1, 0) \cup (0, 1) \cup (1, +\infty) \]