4. Learning Objectives
At the end of this video, you are able to:
1. Recall the difference between function and relation;
2. Recall on the four ways to represent a function; and
3. Evaluate a function.
5. What is the difference between
function and relation?
Function
β’ A function is a relation that is for each
input there is only one output.
β’ A function is a set of ordered pairs
(x,y) such that no two ordered pairs
have the same x-element (abscissa).
β’ A function can be a one-to-one or
many-to-one correspondence but can
not be a one-to-many
correspondence.
Relation
β’ A relation is a set of inputs and
outputs.
β’ A relation is a set of ordered pairs
(x,y).
β’ A relation can be one-to-one, one-to-
many and/or many-to-one
correspondence.
6. Fact or Bluff
1. All functions are relations.
Answer: Fact
2. All relations are functions.
Answer: Bluff
3. A function is a set of ordered pairs
(x,y) such that no two ordered pairs
have the same y-element.
Answer: Bluff
4. A student with a unique
identification number is an
example of a function.
Answer: Fact
5. Not all functions are relations.
Answer: Bluff
7. Four Ways to represent a FUNCTION
Sets of Ordered Pairs
Table of Values
Graphs
Equation
8. Which of the following sets of ordered pairs are functions?
Answer: The sets in a and b are functions.
Multiple Choice
a. h = {(0,0),(1,1),(2,2),(3,3),(4,4),...}
d. H = {(0,0),(0,1),(2,2),(3,3),(4,4),...}
b. g = {(0,0),(1,0),(2,0),(3,0),(4,0),...}
c. f = {(-2,0),(-2,1),(-2,2),(-2,3),(-2,4),...}
9. Which of the following tables below represent a
function?
x 0 1 2 3
y 0 1 2 3
x 1 1 1 1
y -4 -2 0 2
x -1 0 1 2
y 2 2 2 2
x 2 4 2 4
y -2 0 2 6
Answer: A and C
A
B
C
D
10. Which of the following graphs can be graphs of
functions?
Answer: a, b and c
11. Which of the following equations represent a
function?
Answer: a, c and d
π. ππ
+ ππ
= π d. π = ππππ
π. π = πππ β ππ + π c. π = ππ + π
12. Evaluating a function means replacing the variable x in the function
with a given number or expression.
For example, evaluate the function π π₯ = 2π₯ + 4 πππ π₯ = 5.
Just replace the variable βxβ with β5β:
π 5 = 2 5 + 4 = 14
Answer: f(5)=14
Evaluating Functions
13. The function f is define on the real numbers by
π π₯ = 2 + π₯ β π₯2
. What is the value of π β3 ?
Solution:
π π₯ = 2 + π₯ β π₯2, π β3 =?
π β3 = 2 + (β3) β (β3)2 substitute x with -3
π β3 = 2 + β3 β 9 simplify (β32)
π β3 = β10
The value of π β3 is β 10.
14. The function π is define on the real numbers by
g π₯ = (π₯2
+ 1)(3π₯ β 5). What is the value of g 4 ?
Solution:
g π₯ = (π₯2 + 1)(3π₯ β 5), g 4 =?
g 4 = 42 + 1 [3(4) β 5], substitute x with 4
g 4 = (16 + 1)(12 β 5), simplify 42 , multiply 3(4)
g 4 = (17)(7), add (16 + 1), subtract (12 - 5)
g 4 = 119
The value of π 4 is 119.