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High School Mathematics - 2
4.10 Introduction to Functions

Function Definition:
A function is a relation that has exactly one "output" for every "input".

Consider two sets A and B to form the Cartesian product. This Cartesian product forms the relations. Among these relations we select those relations that satisfy the condition - every element on A is related to only one element in set B.
When a relation satisfies this rule we call it function.

Function: Any relation on A x B in which in which every element in set A has a corresponding element on set B.
It is denoted as f: A ->B or f: x -> f(x) or y = f(x) where y is a function of x.
First element is also called domain, abscissa, preimage or the first component.
Second element is also called range, ordinate, image or second component.
A function can be represented by arrow diagram, set-builder notation, Cartesian form.

Representation of a function
Arrow Diagram

Domain = {1,2,3,4}
Range = {D,B,C,A}

Set-builder Notation
f = {(x,y)ly = 3x+5}
Cartesian Form
f = {(1,2), (2,3), (4,5)}


Directions: Solve the following problems. Also write at least 5 examples of your own.
Q 1: Write the domain of h: x-> 2/(x-7)
Domain is all numbers except 7
Domain is set of all natural numbers
Domain is set of all real numbers

Q 2: Which of the following relations are functions?
{(1,2), (1,3), (3,4), (4,5)}
{(1,2), (3,7), (4,-6), (8,11)}
{(2,-3),(3,-4), (3,9), (5,8)}

Q 3: If n(A) = p and n(B) = q, then the possible number of mappings from A to B is n?.
q
p
0

Q 4: If A = {a, b, c} and B = {1,2}then the possible number of mappings from A to B is
8
6
4

Q 5: Let X = {1,2,3,4}. Determine whether r not each relation is a function from X to X.
g = {(3,1), (4,2), (1,1)}
h = {(2,1),(4,4), (3,4)}
f = {(2,3), (1,4), (2,1), (3,2), (4,4)}

Q 6: Given that f(x) = 2x and f: N->N, find the range f .
Answer:

Q 7: Which of the following relations are functions?
y = 3x+2
y < x+3
y > x + 5

Q 8: Write the domain of the function of F: x ->5x , X {0,1,2}
{0, 5, 10}
{0,1,2}
{5, 10, 15}

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Question 10: This question is available to subscribers only!


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