# Find The Domain of Rational Functions - Grade 11 Math Questions With Detailed Solutions

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

 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.

Updated: 20 January 2017 (A Dendane)