The document outlines operations on functions such as addition, subtraction, multiplication, division, and composition. It provides examples of adding, subtracting, multiplying, and dividing fractions and rational expressions. It also defines operations on functions, including addition, subtraction, multiplication, division, and composition, and provides an example of determining various functions based on given functions.

1. OPERATIONS ON
FUNCTIONS

2. LEARNING OUTCOMES:
› At the end of the lesson, the learner is able
to perform addition, subtraction,
multiplication, division, composition of
functions, and solve problems involving
functions.

3. LESSON OUTLINE:
› Review: Operations on algebraic expressions
› Addition, subtraction, multiplication, and
division of functions;
› Function composition

4. ADDITION AND SUBTRACTION:
(a) Find the least common denominator (LCD)
of both fractions.
(b) Rewrite the fractions as equivalent
fractions with the same LCD.
(c) The LCD is the denominator of the
resulting fraction
(d) The sum or difference of the numerators is
the numerator of the resulting fraction.

5. Example 1:
Find the sum of 1/3 and 2/5.
Solution: The LCD of the two fractions is 15.
1
3
+
2
5
=
5
15
+
6
15
=
5+6
15
=
11
15

6. Example 2:
Find the sum of
1
𝑥−3
+
2
𝑥−5
.
Solution: The LCD of the two fractions is (x-
3)(x-5) or 𝑥2
− 8𝑥 + 15.
1
𝑥−3
+
2
𝑥−5
=
𝑥−5
𝑥2−8𝑥+15
+
2 𝑥−3
𝑥2−8𝑥+15
=
𝑥−5+2𝑥−6
𝑥2−8𝑥+15
=
3𝑥−11
𝑥2−8𝑥+15

7. MULTIPLICATION:
(a) Rewrite the numerator and denominator in
terms of its prime factors
(b) Common factors in the numerator and
denominator can be simplified as “1” (this
is often called “cancelling”.
(c) Multiply the numerators together to get
the new numerator.
(d) Multiply the denominators together to get
the new denominator.

8. Example 3:
Find the product of
10
21
𝑎𝑛𝑑
15
8
. Use cancellation of factors
when convenient.
Solution: Express the numerators and denominators of the
two functions into their prime factors. Multiply and cancel
out common factors in the numerator and the denominator
to reduce the final answer to lowest terms.
10
21
⋅
15
8
=
2 ⋅ 5
3 ⋅ 7
⋅
3 ⋅ 5
2 ⋅ 2 ⋅ 2
=
2 ⋅ 5 ⋅ 3 ⋅ 5
3 ⋅ 7 ⋅ 2 ⋅ 2 ⋅ 2
=
25
28

9. Example 4:
Find the product of
𝑥2−4𝑥−5
𝑥2−3𝑥+2
and
𝑥2−5𝑥+6
𝑥2−3𝑥−10
Solution: Express the numerators and denominators of the
two rational expressions into their prime factors. Multiply
and cancel out common factors in the numerator and the
denominator to reduce the final answer to lowest terms.
Note the similarity in the process between this example
and the previous one on fractions.

10. DIVISION:
(a) To divide two fractions or rational
expressions, multiply the dividend with the
reciprocal of the divisor.

11. Example 5:
Divide
2𝑥2+𝑥−6
2𝑥2+7𝑥+5
by
𝑥2−2𝑥−8
2𝑥2−3𝑥−20
Solution:

12. DEFINITION
Let f & g be functions.
1. Their sum, denoted by f + g, is the function denoted
by (f+g)(x) = f(x) + g(x).
2. Their difference, denoted by f – g, is the function
denoted by (f-g)(x) = f(x) – g(x).
3. Their product, denoted by 𝑓 ⋅ 𝑔 𝑥 = 𝑓 𝑥 ⋅ 𝑔 𝑥
4. Their quotient, denoted by 𝑓
𝑔, is the function denoted
by 𝑓
𝑔 𝑥 = 𝑓 𝑥 𝑔 𝑥 , excluding the values of x
where 𝑔 𝑥 = 0

13. Use the following functions below for example 1:
› 𝑓 𝑥 = 𝑥 + 3
› p 𝑥 = 2𝑥 − 7
› v 𝑥 = 𝑥2 + 5𝑥 + 4
› 𝑔 𝑥 = 𝑥2 + 2𝑥 − 8
› ℎ 𝑥 =
𝑥+7
2−𝑥
› 𝑡 𝑥 =
𝑥−2
𝑥+3

14. Example 1: determine the ff. functions
(a) 𝑣 + 𝑔 𝑥
(b) 𝑓 ⋅ 𝑝 𝑥
(c) 𝑓 + ℎ 𝑥
(d) 𝑝 − 𝑓 𝑥
(e) 𝑣
𝑔 𝑥

15. Solution.
(a) 𝑣 + 𝑔 𝑥
= 𝑥2
+ 5𝑥 + 4 + 𝑥2
+ 2𝑥 − 8
= 𝑥2
+ 5𝑥 + 4 + 𝑥2
+ 2𝑥 − 8
= 2𝑥2
+ 7𝑥 − 4

16. Solution:
(b) 𝑓 ⋅ 𝑝 𝑥
= 𝑥 + 3 2𝑥 − 7
= 2𝑥2
− 𝑥 − 21

17. Solution:
(c) 𝑓 + ℎ 𝑥
= 𝑥 + 3 +
𝑥+7
2−𝑥
= 𝑥 + 3 ⋅
2−𝑥
2−𝑥
+
𝑥+7
2−𝑥
=
𝑥+3 2−𝑥 + 𝑥+7
2−𝑥
=
6−𝑥−𝑥2+𝑥+7
2−𝑥
=
𝑥2−13
2−𝑥

18. Solution:
(d) 𝑝 − 𝑓 𝑥
= 2𝑥 − 7 − 𝑥 + 3
= 2𝑥 − 7 − 𝑥 − 3
= 𝑥 − 10

19. Solution:
(e) 𝑣
𝑔 𝑥
= 𝑥2 + 5𝑥 + 4 ÷ 𝑥2 + 2𝑥 − 8
=
𝑥2+5𝑥+4
𝑥2+2𝑥−8