Lesson 3
Domain and Range
of Rational Function


1

Domain of a function
the set of all values of π‘₯ that have corresponding values of 𝑦; it contains all values that go into the function

$$Example:𝑓(π‘₯)=\frac{5}{π‘₯}$$

note: scrolable

a. Make sure that the function is in its simplest form. The numerator and the denominator of 𝑓(π‘₯) have no common factor, thus the function is already in its simplest form.
b. Find the vertical asymptote(s) of 𝑓(π‘₯) by setting the denominator equal to zero. Then, solve the resulting equation.
x=0
c. Exclude the vertical asymptote(s) obtained in the previous step from the set of real numbers. The remaining elements of the set of real numbers comprise the domain of 𝑓 π‘₯.


1

Domain of a function
the set of all values of π‘₯ that have corresponding values of 𝑦; it contains all values that go into the function

Example:

Thus, the domain of \(f(x)=\frac{5}{x}\) is the set of all real numbers \(x\) such that \(x \neq 0\). In symbols, \(D: \{x|x \neq 0\}\).


2

Range of a function
the set of all values of 𝑦 that can be obtained from the possible values of π‘₯; it contains all possible values of the function

Example:

Consider the rational function\(f(x)=\frac{5}{x}\). To find the range of \(f(x)\), we can do the following steps.


2

Range of a function
the set of all values of 𝑦 that can be obtained from the possible values of π‘₯; it contains all possible values of the function

$$Example:𝑓(π‘₯)=\frac{5}{π‘₯}$$

note: scrolable

a. Make sure that the function is in its simplest form.
b. Find the horizontal asymptote of 𝑓(π‘₯) by comparing the degrees of the numerator and the denominator. The degree of the numerator (0) is less than the degree of the denominator (1). Recall that if 𝑛 < π‘š, then the horizontal asymptote of the function is the line
y = 0.
b. The degree of the numerator (0) is less than the degree of the denominator (1). Recall that if 𝑛 < π‘š, then the horizontal asymptote of the function is the line
𝑦 = 0.
c. Exclude the horizontal asymptote obtained in the previous step from the set of real numbers. The remaining elements of the set of real numbers comprise the range of 𝑓(π‘₯).


2

Range of a function
the set of all values of 𝑦 that can be obtained from the possible values of π‘₯; it contains all possible values of the function

Example:

Thus, the domain of \(f(x)=\frac{5}{x}\) is the set of all real numbers \(x\) such that \(x \neq 0\). In symbols, \(D: \{x|x \neq 0\}\).


Example 1: Determine the domain and range of \(f(x)=\frac{π‘₯+4}{xβˆ’7}.\)
Solution: To find the domain of \(f(x)=\frac{π‘₯+4}{xβˆ’7}.\)

First, find the vertical asymptote(s) of 𝑓(π‘₯) by setting the denominator equal to zero and solving the resulting equation.

x - 7 = 0
x = 7


Example 1: Determine the domain and range of \(f(x)=\frac{π‘₯+4}{xβˆ’7}.\)
Solution: To find the domain of \(f(x)=\frac{π‘₯+4}{xβˆ’7}.\)

Then, exclude the vertical asymptote(s) obtained in the previous step from the set of real numbers. The remaining elements of the set of real numbers comprise the domain of 𝑓(π‘₯).


Example 1: Determine the domain and range of \(f(x)=\frac{π‘₯+4}{xβˆ’7}.\)
Solution:

Thus, the domain of \(f(x)=\frac{π‘₯+4}{xβˆ’7}.\) is the set of all real numbers π‘₯ such that π‘₯ β‰  7. In symbols, \(D: \{x|x \neq 7\}\).


Example 1: Determine the domain and range of \(f(x)=\frac{π‘₯+4}{xβˆ’7}.\)
Solution: To find the range of \(f(x)=\frac{π‘₯+4}{xβˆ’7}.\)

First, find the horizontal asymptote of 𝑓(π‘₯) by comparing the degrees of the numerator and the denominator.


Example 1: Determine the domain and range of \(f(x)=\frac{π‘₯+4}{xβˆ’7}.\)
Solution: To find the range of \(f(x)=\frac{π‘₯+4}{xβˆ’7}.\)

The numerator and the denominator of \(f(x)\) have the same degree (1). Recall that if \(n = m\), then the horizontal asymptote of the function is the line \(y = \frac{a_n}{b_m}\), where the expressions \(a_n\) and \(b_m\) are the leading coefficients of \(P(x)\) and \(Q(x)\), respectively. For the given rational function, this expression is equal to 1.


Example 1: Determine the domain and range of \(f(x)=\frac{π‘₯+4}{xβˆ’7}.\)
Solution: To find the range of \(f(x)=\frac{π‘₯+4}{xβˆ’7}.\)

Then, exclude the horizontal asymptote obtained in the previous step from the set of real numbers. The remaining elements of the set of real numbers comprise the range of 𝑓(π‘₯).


Example 1: Determine the domain and range of \(f(x)=\frac{π‘₯+4}{xβˆ’7}.\)
Solution: To find the range of \(f(x)=\frac{π‘₯+4}{xβˆ’7}.\)

Thus, the range of \(f(x)=\frac{π‘₯+4}{xβˆ’7}.\) is the set of all real numbers 𝑦such that 𝑦 β‰  1. In symbols, \(R: \{y|y \neq 1\}\).


Example 2: Find the domain and range of \(f(x)=\frac{(π‘₯+2)(x-3}{(x+2)(x-5)}.\)
Solution: First, reduce the rational function to its simplest form.

The numerator and the denominator of 𝑓(π‘₯) have a common factor of π‘₯ + 2. Recall that when the numerator and the denominator have common factor(s), hole(s) at the zero(s) is(are) produced. Hence, the simplified form of the function is \(f(x)=\frac{π‘₯-3}{xβˆ’5}.\)


Example 2: Find the domain and range of \(f(x)=\frac{(π‘₯+2)(x-3}{(x+2)(x-5)}.\)
Solution: To find the domain of \(f(x)=\frac{π‘₯-3}{xβˆ’5}.\)

First, find the vertical asymptote(s) of 𝑓(π‘₯) by setting the denominator equal to zero and solving the resulting equation.

π‘₯ βˆ’ 5 = 0
π‘₯ = 5


Example 2: Find the domain and range of \(f(x)=\frac{(π‘₯+2)(x-3}{(x+2)(x-5)}.\)
Solution: To find the domain of \(f(x)=\frac{π‘₯-3}{xβˆ’5}.\)

Then, exclude the vertical asymptote(s) obtained in the previous step from the set of real numbers. Reject the zero of the common factor that is canceled.
Equating the common factor π‘₯ + 2 to zero yields to π‘₯ = βˆ’2. This is not included in the domain of 𝑓(π‘₯). The remaining elements of the set of real numbers comprise the domain of π‘₯ = 5


Example 2: Find the domain and range of \(f(x)=\frac{(π‘₯+2)(x-3}{(x+2)(x-5)}.\)
Solution: To find the domain of \(f(x)=\frac{π‘₯-3}{xβˆ’5}.\)

Equating the common factor π‘₯ + 2 to zero yields to π‘₯ = βˆ’2. This is not included in the domain of 𝑓(π‘₯). The remaining elements of the set of real numbers comprise the domain of 𝑓(π‘₯).


Example 2: Find the domain and range of \(f(x)=\frac{(π‘₯+2)(x-3}{(x+2)(x-5)}.\)
Solution:

Thus, the domain of \(f(x)=\frac{(π‘₯+2)(x-3}{(x+2)(x-5)}.\) is the set of all real numbers π‘₯ such that π‘₯ β‰  βˆ’2, 5. In symbols, \(D: \{x|x \neq -2, 5\}\).


Example 2: Find the domain and range of \(f(x)=\frac{(π‘₯+2)(x-3}{(x+2)(x-5)}.\)
Solution: To find the range of \(f(x)=\frac{π‘₯-3}{xβˆ’5}.\)

First, find the horizontal asymptote of 𝑓(π‘₯) by comparing the degrees of the numerator and the denominator.


Example 2: Find the domain and range of \(f(x)=\frac{(π‘₯+2)(x-3}{(x+2)(x-5)}.\)
Solution: To find the range of \(f(x)=\frac{π‘₯-3}{xβˆ’5}.\)

The numerator and the denominator of \(f(x)\) have the same degree (1). Recall that if \(n = m\), then the horizontal asymptote of the function is the line \(y = \frac{a_n}{b_m}\), where the expressions \(a_n\) and \(b_m\) are the leading coefficients of \(P(x)\) and \(Q(x)\), respectively. For the given rational function, this expression is equal to 1.


Example 2: Find the domain and range of \(f(x)=\frac{(π‘₯+2)(x-3}{(x+2)(x-5)}.\)
Solution: To find the range of \(f(x)=\frac{π‘₯-3}{xβˆ’5}.\)

Then, exclude the horizontal asymptote obtained in the previous step from the set of real numbers. If there is a common factor canceled, say (π‘₯ βˆ’ π‘˜), reject the value for 𝑓(π‘˜). The remaining elements of the set of real numbers comprise the range of 𝑓(π‘₯).


Example 2: Find the domain and range of \(f(x)=\frac{(π‘₯+2)(x-3}{(x+2)(x-5)}.\)
Solution: To find the range of \(f(x)=\frac{π‘₯-3}{xβˆ’5}.\)

The common factor canceled is \(x + 2\). Hence, we reject the value for \(f(-2)\).


Example 2: Find the domain and range of \(f(x)=\frac{(π‘₯+2)(x-3}{(x+2)(x-5)}.\)
Solution: To find the range of \(f(x)=\frac{π‘₯-3}{xβˆ’5}.\)

\(f(x) = \frac{x - 3}{x - 5}\)
\(f(-2) = \frac{-2 - 3}{-2 - 5}\)
\(f(-2) = \frac{-5}{-7}\)
\(f(-2) = \frac{5}{7}\)


Example 2: Find the domain and range of \(f(x)=\frac{(π‘₯+2)(x-3}{(x+2)(x-5)}.\)
Solution:

Thus, the range of \(f(x) = \frac{(x+2)(x-3)}{(x+2)(x-5)}\) is the set of all real numbers \(y\) such that \(y \neq \frac{5}{7}\),1. In symbols, \(R: \{y \mid y \neq \frac{5}{7}, 1\}\).


πŸ”₯Group-xXJLDXxπŸ”₯

11-ICT

Techer:
Sir. Jervin Sta Monica

Desmond M. Torres,
Alija Raine Manjares,
Elzen Richohermoso,
Taynie Gacita,
Cleanuar Shan Stefan.