1. Lesson 3.8
Functions

2. No input value is repeated so the relation is a function
Definition
A function is a special relation between values: The input value (domain) gives
back exactly one unique Output value (range).
Class Grade Function
Student
Grade 80 85 96
Bilal
Peter
Selma
Ahmad
George
{(B, 85), (P, 96), (S, 80), (A, 85), (G, 96)}
No input value is repeated so the relation is a function

3. Example of a relation that is not a function
Has visited
Person
Country
Paris
London
Dubai
New York
Ahmed
Rami
Sally
Nancy
{(A, D), (A, P), (R, N), (S, P), (S, L), (S, D), (N, D)}
The inputs A and S are repeated so the relation is not a function

4. x y F(x) = x2 – 1 { } { } Function Notation (x2 – 1) function Domain
{ }
{ }
Domain 2, 5, 7, 10
Range 3,
24,
48, 99
F(x) = x2 – 1
Output
input
Function name

5. F: x 3 – 2x; D = {0, 1, 2, 3} f(x) = 3 – 2x R = {3, 1, -1, -3} f(0) =
Example
The Domain D and the rule of some function are given. Find the range
F: x – 2x;
D = {0, 1, 2, 3}
f(x) = 3 – 2x
f(0) =
3 – 2(0) = 3
f(1) =
3 – 2(1) = 1
f(2) =
3 – 2(2) = -1
f(3) =
3 – 2(3) = -3
R = {3, 1, -1, -3}

6. 1) F(x) = x3 – 3 2) F(x) = Domain of the function
The domain of a function is all the values that an input is allowed to take on.
Give the domain of each function
1) F(x) = x3 – 3
There are no values that I can't plug in for x. when I have a polynomial, the answer is always that the domain is ”all real numbers”
2) F(x) =
The only values that x can not take on are those which would cause division by zero.
Here x can not take (1), so my domain will be “All real numbers except 1”

7. 3) F(x) = x – 2 ≥ 0 x ≥ 2 Domain of the function
The only problem I have with this function is that I cannot have a negative inside the
square root. So I'll set the insides ≥ 0, and solve. The result will be my domain:
x – 2 ≥ 0
x ≥ 2
Then the domain is “all x ≥ 2”

8. f(g(1)) f(g(1)) Composite functions
Consider the functions:
f(x) = 3x2 + 2
and
g(x) = x + 1
is a composite function, where g is performed first and then f is performed
on the result of g. The function fg may be found using a flow diagram
Examples: find the indicated value: f(g(1))
f(g(1))
g(x)
x + 1
f(x)
3x2 + 2
Method 1 1
input 2
input fg
= 14
f(g(1))
Method 2
Step1: find g(1)
= (1) + 1 = 2
Step2: find f(2)
= 3(2)2 + 2 = = 14

9. H.W
(4, 8, 12) + (30 – 40) Even
page 144, 145