How to Find the Domain of Rational Functions?

A step by step tutorial, with detailed solutions, on how to find the domain of rational functions is presented. Problems matched to the exercises with solutions at the bottom of the page are also presented.

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.
also a Step by Step Calculator to Find Domain of a Function is included.

Find the Domain of a Rational Function: Examples with Solutions

Example 1

Find the domain of the function \( f \) defined by \[ f(x) = \dfrac{1}{x-2} \] Solution to Example 1
\( f(x) \) 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 above inequality for \( x \) to obtain the domain: \( x \neq 2 \)
Which in interval form may be written as follows
\((- \infty , 2) \cup (2 , +\infty)\)

Matched Problem 1

Find the domain of the function \( f \) given by \[ f(x) = \dfrac{1}{6-x} \]

Example 2

Find the domain of the function \( f \) defined by \[ f(x) = \dfrac{x+3}{x^2+7} \] Solution to Example 2
For \( f(x) \) to have real values, the denominator must be different from zero. Hence
\( x^2 + 7 \neq 0 \)
Expression \( x^2 + 7 \) is always positive (square added to a positive number). Hence the domain of \( f \) is given by the interval
\((- \infty , +\infty)\)

Matched Problem 2

Find the domain of function \( f \) defined by: \[ f(x) = \dfrac{2x+4}{-x^2-9} \]

Example 3

Find the domain of the function \( f \) defined by: \[ f(x) = \dfrac{2x+9}{2x^2+x-15} \]
Solution to Example 3
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
\( 2 x^2 + x - 15 = 0 \)
The roots are
\( -3 \) and \( \frac{5}{2} \)
The denominator \( 2x^2 + x - 15 \) is not equal to zeros for all real values except \( -3 \) and \( \frac{5}{2} \). Hence the domain of the given function is given by
\((- \infty , - 3) \cup (- 3 , \frac{5}{2}) \cup (\frac{5}{2} , +\infty)\)

Matched Problem 3

Find the domain of function \( f \) defined by: \[ f(x) = \dfrac{-x+4}{-3x^2+2x+8} \]

Example 4

Find the domain of the function \( f \) given by: \[ f(x) = \dfrac{2}{2x-6} - \dfrac{x}{4x+7} \] Solution to Example 4
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 \( x = 3 \)
\( 4x + 7 \) gives \( x = - \frac{7}{4} \)
\( f(x) \) is real for all real values except \( 3 \) and \( - \frac{7}{4} \). The domain of the above function is given by
\((- \infty , -\frac{7}{4}) \cup (-\frac{7}{4} , 3) \cup (3 , +\infty)\)

Matched Problem 4

Find the domain of the function \( f \) given by: \[ f(x) = \dfrac{2 - 7x}{-x-4} - \dfrac{x+1}{-7x+3} \]

Example 5

Find the domain of the function \( f \) defined by: \[ f(x) = \dfrac{-5x+6}{x^3 - x} \]
Solution to Example 5
Let us first find the values that makes the denominator equal to zero
\( x^3 - x = 0 \)
Factor the expression on the left hand side of the equation
\( x(x^2 - 1) = 0 \)
\( x(x + 1)(x - 1) = 0 \)
Solve the above equation to obtain the solution set as follows:
\( x = 0 \), \( x = 1 \) and \( x = -1 \)
The domain is given in interval form as follows:
\((- \infty , - 1) \cup (- 1 , 0) \cup (0 , 1) \cup (1 , +\infty)\)

Matched Problem 5

Find the domain of function \( f \) defined by: \[ f(x) = \dfrac{-7x-2}{x^3 +2x^2-3x} \]

Answers to Matched Problems

\((- \infty , 6) \cup (6 , + \infty)\)
\((- \infty , + \infty)\)
\((- \infty , - 4 / 3) \cup (- 4 / 3 , 2) \cup (2 , \infty)\)
\((- \infty , - 4) \cup (- 4 , 3 / 7) \cup (3 / 7 , \infty)\)
\((- \infty , - 3) \cup (- 3 , 0) \cup (0 , 1) \cup (1 , \infty)\)