Find The Domain of Rational Functions Questions With Solutions

How to find the domain of rational functions? Examples are presented along with detailed solutions and explanations and also graphical interpretation.

Division by Zero and 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: $$(-\infty , 0) \cup (0 , \infty)$$

Examples with Solutions

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.