How to find the domain of a rational function?

We first need to understand that \( \dfrac{1}{x} \) takes real values only if the denominator is not equal to 0. In this case \( x \ne 0 \). This can easily be verified by examining the graph of \( y = \dfrac{1}{x} \) shown below: The graph "exists" for all values of \( x \) except 0.

The domain in interval form is given by: \( (-\infty , 0) \cup (0 , \infty) \)

Example 1: Find the domain of the function \( f(x) = \dfrac{1}{x - 2} \).

Solution:

The given function takes real values and is therefore real if

\( x - 2 \ne 0 \) or \( x \ne 2 \)

In interval form, the domain is given by

\( (-\infty , 2) \cup (2 , \infty) \)

Below is shown the graph of \( f \) and we can see that the given function is undefined at x = 2.

Example 2: Find the domain of the function \( f(x) = \dfrac{x + 3}{(x - 1)(x + 2)} \).

Solution:

The given function takes real values and is therefore real if

\( (x - 1)(x + 2) \ne 0 \)

The expression (x - 1)(x + 2) is not equal to zero if x ≠ 1 and x ≠ -2.

In interval form, the domain is given by

\( (-\infty , - 2) \cup (-2 ,1 ) \cup (1 , \infty) \)

Below is shown the graph of \( f \) and we can see that the given function is undefined at x = -2 and x = 1.

Example 3: Find the domain of the function \( f(x) = \dfrac{1}{ {x^2 - 4}} \).

Solution:

The given function takes real values and is therefore real if

\( x^2 - 4 \ne 0 \)

Factor x^{ 2} - 4 and rewrite the inequality as

\( (x - 2)(x + 2) \ne 0 \)

The expression (x - 2)(x + 2) is not equal to zero if x ≠ -2 and x ≠ 2.

In interval form, the domain is given by

\( (-\infty , - 2) \cup (-2 ,2 ) \cup (2 , \infty) \)

Below is shown the graph of \( f \) and we can see that the given function is undefined at x = -2 and x = 2.

Example 4: Find the domain of the function \( f(x) = \dfrac{x + 1}{x^2 + x - 2} \).

Solution:

The given function takes real values and is therefore real if

\( x^2 + x - 2 \ne 0 \)

Factor x^{ 2} + x - 2 and rewrite the inequality as

\( (x - 1)(x + 2) \ne 0 \)

The expression (x - 1)(x + 2) is not equal to zero if x ≠ 1 and x ≠ - 2.

In interval form, the domain is given by

\( (-\infty , - 2) \cup (-2 ,1 ) \cup (1 , \infty) \)

Below is shown the graph of \( f \) and we can see that the given function is undefined at x = -2 and x = 1.

Example 5: Find the domain of the function \( f(x) = \dfrac{x^2 - 1}{x^2 + x + 5} \).

Solution:

The given function takes real values and is therefore real if

\( x^2 + x + 5 \ne 0 \)

The expression x^{ 2} + x + 5 cannot be factored over the reals. We therefore need to solve the quadratic equation using the discriminant \( \Delta \).

\( x^2 + x + 5 = 0 \)

\( \Delta = b^2 - 4 a c = (1)^2 - 4(1)(5) = -19 \)

The discriminant is negative and therefore no real value of x makes the expression x^{ 2} + x + 5 equal to zero. The domain is the set of all real numbers

In interval form, the domain is given by

\( (-\infty , \infty) \)

Below is shown the graph of \( f \) and we can see that it defined for all values of x real.