Lesson 3 Operation on Functions

LESSON 3
OPERATIONS ON
FUNCTIONS

• In this lesson, the different operations on functions
are to be discussed. These are the Sum, the
Difference, the Product, the Quotient, and the
Product of a Constant and a Function.
• The process on how we perform the different
operations on algebraic expressions are to applied
in order to perform the said operations.

LESSON OBJECTIVES:
At the end of the lesson, the student are expected to:
1. identify the different operations on functions.
2. enumerate the steps on how to perform the different
operations on functions.
3. perform the indicated operations of the given
functions; and
4. manifest the value of patience, and analytical and
critical thinking skills.

THE SUM OF TWO FUNCTIONS
Definition. Given two functions f and g,
the sum of two functions, f + g, is
defined by the equation,
(f + g) (x) = f(x) + g(x)

Illustrative examples:
Find the sum of the given functions given
the following:
Given f(x) = 6x – 1.
g(x) = x + 5
P(x) = x² + 3x + 2
h(x) = x²- 5x + 4

a.
f + g
Equation:
(f + g) (x) = f(x) + g(x)
Substitute the given functions to the equation:
(f + g) (x) = 6x – 1 +x + 5
Simplify:
(f + g) (x) = 7x + 4
Therefore, the sum of the functions f and g is 7x + 4.

B.
g + P
Equation:
(g + P) (x) = g(x) + P(x)
Solution the given functions to the equation:
(g + P) (x) = x + 5 + x²+ 3x + 2
Simplify:
(g + P) (x) = x² + 4x + 7
Therefore, the sum of the functions g and P is x² + 4x + 7.

c.
f + h
Equation:
(f + h) (x) = f(x) + h(x)
(f + h) = 6x – 1 + x² - 5x + 4
Simplify:
(f + h) (x) = x² + x + 3
Therefore, the sum of the functions f and h is x² + x + 3.

Remember:
In finding the sum of the functions, the following
steps will be applied:
1. Write the equation based on the given
functions.
2. Substitute the given values of the functions to
the equation.
3. Simplify the values to get the final answer.

THE DIFFERENCE OF TWO FUNCTIONS
Definition: Given two functions, f and g
the difference of two functions, f – g, is
(f – g) (x) = f(x) – g(x)

Illustrative Examples:
Find the difference of the given functions
given the following:
Given:
f(x) = 6x – 1
g(x) = x + 5
P(x) = x² + 3x + 2
h(x) =x² - 5x + 4

a.
f – g
Equation:
(f – g) (x) = f(x) –g(x)
(f – g) (x) = 6x – 1 – (x + 5)
Distribute the negative sign to the quantities in the parenthesis:
(f – g) (x) = 6x – 1 – x – 5
Simplify:
(f – g) (x) = 5x – 6
Therefore, the difference of the functions f and g is 5x – 6.

B
g – P
Equation:
(g – P) (x) = g(x) – P(x)
(g – P) (x) = x + 5 – (x² + 3x + 2)
Distribute the negative sign to the quantities in the parenthesis:
(g – P) (x) = x + 5 – x² – 3x – 2
Simplify:
(g – P) (x) = -x² – 2x + 3
Therefore, the difference of the functions g and P is –x² – 2x + 3.

c.
f – h
Equation:
(f – h) (x) = f(x) – h(x)
(f – h) (x) = 6x – 1 – (x² -5x + 4)
Distribute the negative signs to the quantities in the parenthesis:
(f – h) (x) = 6x – 1 – x² + 5x – 4
Simplify:
(f – h) (x) = -x² + 11x – 5
Therefore, the difference of the functions f and h is –x² + 11x – 5.

Remember:
In finding the difference of two functions, the
following steps will be applied:
1. write the equation based on the given functions.
2. substitute the given values of the functions to the
equation.
3. distribute the negative or minus sign on the
quantities inside the parenthesis.
4. simplify the values to get the final answer.

THE PRODUCT OF TWO FUNCTIONS
Definition: Given two functions, f and g,
the product of two functions, f and g, is
(f . g) (x) = f(x) . g(x)

Illustrative examples:
Find the products of the following functions given:
Given:
f(x) = x – 1
g(x) = x + 5
P(x) = x² + 3x + 2
h(x) = x² – 5x + 4
G(x) = 2x

a.
G. f
Equation:
(G. f) (x) = G(x) . f(x)
(G . f) (x) = 2x (x – 1)
Multiply the quantities by applying the Law of Exponents:
(G . f) (x) = 2x² – 2x
Therefore, the product of the functions G and f is 2x² – 2x.

b.
f . g
Equation:
(f . g) (x) = f(x) . g(x)
Substitute the given functions to the equations:
(f . g) (x) = (x – 1) (x + 5)
Multiply the quantities by applying the Laws of Exponents
horizontally or vertically:
(f . g) (x) = x² + 5x – x – 5
Simplify
(f . g) (x) = x² + 4x – 5
Therefore, the product of the functions g and h is x² + 4x – 5.

c.
g and h
Equation:
(g . h) (x) = g(x) . h(x)
(g . h) (x) = (x + 5) (x² -5x + 4)

Multiply the quantities by applying the Laws
of Exponents horizontally or vertically:
(g . h) (x) = x³ – 5x² + 4x + 5x² – 25x + 20
Simplify
(g . h) (x) = x³ – 21x + 20
Therefore, the product of the functions g
and h is x³ – 21x + 20.

d.
P . h
Equation:
(P . h) (x) = P(x) . h(x)
(P . h) (x) = (x² + 3x + 2) (x² – 5x + 4)

Multiply the quantities by applying the Laws of
Exponents horizontally or vertically:
(P . h) (x) = x⁴ – 5x³ + 4x² + 3x – 15x² + 12x +
2x – 10x + 8x – 2x² – 9x² = 2x = 8
Simplify:
(P . h) (x) = x⁴ – 2x³ – 9x² + 2x + 8
Therefore, the product of the functions P and
h is x⁴ – 2x³ – 9x² + 2x + 8.

Remember
In finding the product of two functions, the following
steps will be applied:
equation.
3. Multiply the given functions horizontally and vertically by
applying the laws of exponents.
4. simplify the values to get the final answer.

THE QUOTIENT OF TWO FUNCTIONS
Definition: Given two functions, f and g,
the quotient of two functions,
f/g, is defined y the equation,
(f/g) (x) = f(x)/g(x)

illustrative Examples:
Given: f(x) = x – 1
g(x) = x + 5
P(x) = x² – 2x + 1
h(x) = x² + 10x + 25
G(x) = 2x – 2

a.
f/g
Equation:
(f/g) (x) = f(x) / g(x)
(f/g) (x) = x – 1 / x – 5
Since x – 1 is not divisible by x + 5, the quotient of
f and g is x – 1 / x + 5

b.
G/f
Equation:
(G/f) (x) = G(x) / f(x)
Substitute the given functions to the
equation:
(G/f) (x) = 2x – 2 / x – 1

Divide the numerator by the
denominator or apply the different kinds of
factoring:
(G/f) (x) = 2(x – 1) / x – 1
Simplify:
(G/f) (x) = 2
Therefore the quotient of the functions
G and f is 2.

c.
g/h
Equation:
(g/h) (x) = g(x) / h(x)
Substitute the given functions to the
equation
(g/h) (x) = x + 5 / x + 10x + 25

Divide the numerator by the denominator
or apply the different kinds of factoring:
(g/h) (x) = x + 5 / ( x + 5 ) ( x + 5 )
Simplify:
(g/h) (x) = 1 / x + 5
Therefore the quotient of the functions
g and h is 1 / x + 5.

Remember:
In finding the quotient of two functions, the
equation.
3. divide the numerator by the denominator or apply
the different kinds of factoring.
4. simplify the value to get the final answer.

THE PRODUCT OF A CONSTANT
AND A FUNCTION
Definition: Given the constant c and the
function f, the product of a constant
and a function, (c . f) (x),is defined by
the equation,
(c . f) (x) = c . f(x)

Illustrative Example:
Find the sum of the given functions given the following:
Given:
f(x) = 6x – 1
P(x) = x² + 3x + 2
h(x) = x² – 5x + 4
g(x) = x + 5

a.
4g
Equation:
(4 . g) (x) = 4 . g (x)
(4 . g) (x) – 4 (x + 5)
Distribute the constant to the quantities inside the parenthesis
and simplify:
(4 . g) (x) = 4x + 20
Therefore, the product of the constant 4 and the function g is 4x
+ 20.

b.
-2P
Equation:
(-2 . P) (x) = -2 . P (x)
(-2 . p) (x) = -2 (x² + 3x + 2)
Distribute the constant to the quantities inside the parenthesis
and simplify:
(-2 . P) (x) = -2x² – 6x – 4
Therefore, the product of the constant -2 and the function P is -
2x² – 6x – 4.

c.
-3h
Equation:
(-3 . h) (x) = -3 . h (x)
(-3 . h) (x) = -3 (x² – 5x + 4)
Distribute the constant to the quantities inside the
parenthesis and simplify:
(-3 . h) (x) = 3x² + 15x – 12
Therefore, the product of the constant -3 and the
function h is (-3 . h) (x) = -3x² + 15x – 12.

Remember:
In finding the product of a constant and a function, the
equation.
3. distribute the constant to the quantities in the parenthesis
and simplify.

Lesson 3 Operation on Functions

Lesson 3 Operation on Functions

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