AA Section 1-3

Section 1-3Function Notations

Warm-up1. If , find y when x = 0.

y = 3x − 2 2. If ,find y when x = 3.

y = 4x + x2

16

3. If ,find y when x = 5.

y =4 − xx2


y = 3x −1


y = 3x − 2 2. If ,find y when x = 3.

y = 4x + x2

16


y =4 − xx2


y = 3x −1

y = 3(0) − 2


y = 3x − 2 2. If ,find y when x = 3.

y = 4x + x2

16


y =4 − xx2


y = 3x −1

y = 3(0) − 2 y = 0 − 2


y = 3x − 2 2. If ,find y when x = 3.

y = 4x + x2

16


y =4 − xx2


y = 3x −1

y = 3(0) − 2 y = 0 − 2

y = −2


y = 3x − 2 2. If ,find y when x = 3.

y = 4x + x2

16


y =4 − xx2


y = 3x −1

y = 3(0) − 2 y = 0 − 2

y = −2

y = 4(3) + 32

16


y = 3x − 2 2. If ,find y when x = 3.

y = 4x + x2

16


y =4 − xx2


y = 3x −1

y = 3(0) − 2 y = 0 − 2

y = −2

y = 4(3) + 32

16 y = 12 + 916


y = 3x − 2 2. If ,find y when x = 3.

y = 4x + x2

16


y =4 − xx2


y = 3x −1

y = 3(0) − 2 y = 0 − 2

y = −2

y = 4(3) + 32

16 y = 12 + 916

y = 12 916


y = 3x − 2 2. If ,find y when x = 3.

y = 4x + x2

16


y =4 − xx2


y = 3x −1

y = 3(0) − 2 y = 0 − 2

y = −2

y = 4(3) + 32

16 y = 12 + 916

y = 12 916 =

20116


y = 3x − 2 2. If ,find y when x = 3.

y = 4x + x2

16


y =4 − xx2


y = 3x −1

y = 3(0) − 2 y = 0 − 2

y = −2

y = 4(3) + 32

16 y = 12 + 916

y = 12 916 =

20116

y =

4 − 552


y = 3x − 2 2. If ,find y when x = 3.

y = 4x + x2

16


y =4 − xx2


y = 3x −1

y = 3(0) − 2 y = 0 − 2

y = −2

y = 4(3) + 32

16 y = 12 + 916

y = 12 916 =

20116

y =

4 − 552

y =−125


y = 3x − 2 2. If ,find y when x = 3.

y = 4x + x2

16


y =4 − xx2


y = 3x −1

y = 3(0) − 2 y = 0 − 2

y = −2

y = 4(3) + 32

16 y = 12 + 916

y = 12 916 =

20116

y =

4 − 552

y =−125

y = 36−1


y = 3x − 2 2. If ,find y when x = 3.

y = 4x + x2

16


y =4 − xx2


y = 3x −1

y = 3(0) − 2 y = 0 − 2

y = −2

y = 4(3) + 32

16 y = 12 + 916

y = 12 916 =

20116

y =

4 − 552

y =−125

y = 36−1 y = 35


y = 3x − 2 2. If ,find y when x = 3.

y = 4x + x2

16


y =4 − xx2


y = 3x −1

y = 3(0) − 2 y = 0 − 2

y = −2

y = 4(3) + 32

16 y = 12 + 916

y = 12 916 =

20116

y =

4 − 552

y =−125

y = 36−1 y = 35

y = 243

Euler Notation:

Euler Notation: f(x), read as “f of x”

Euler Notation: f(x), read as “f of x”The variable inside the parentheses is the independent variable, while the entire “f(x)” is the dependent variable. This means y=f(x).


Argument:


Argument: Yet another word to represent the independent variable; can be filled with anything that is in the possible domain



Value:



Value: The dependent variable

Example 1 P(x) = x3,Q(x) = 3x + 3,R(x) = x

2x − 4Evaluate the following:

a. P(4) c. R(2) b. Q(−3)



a. P(4) c. R(2)

P(4) = 43 b. Q(−3)



a. P(4) c. R(2)

P(4) = 43

P(4) = 64

b. Q(−3)



a. P(4) c. R(2)

P(4) = 43

P(4) = 64 Q(−3) = 3(−3) + 3

b. Q(−3)



a. P(4) c. R(2)

P(4) = 43

P(4) = 64 Q(−3) = −9 + 3 Q(−3) = 3(−3) + 3

b. Q(−3)



a. P(4) c. R(2)

P(4) = 43

P(4) = 64

Q(−3) = −6 Q(−3) = −9 + 3

Q(−3) = 3(−3) + 3 b. Q(−3)



a. P(4) c. R(2)

P(4) = 43

P(4) = 64

Q(−3) = −6 Q(−3) = −9 + 3

Q(−3) = 3(−3) + 3 b. Q(−3)

R(2) = 2

2(2) − 4



a. P(4) c. R(2)

P(4) = 43

P(4) = 64

Q(−3) = −6 Q(−3) = −9 + 3

Q(−3) = 3(−3) + 3 b. Q(−3)

R(2) = 2

2(2) − 4

R(2) = 2

4 − 4



a. P(4) c. R(2)

P(4) = 43

P(4) = 64

Q(−3) = −6 Q(−3) = −9 + 3

Q(−3) = 3(−3) + 3 b. Q(−3)

R(2) = 2

2(2) − 4

R(2) = 2

4 − 4

R(2) = 2

0



a. P(4) c. R(2)

P(4) = 43

P(4) = 64

Q(−3) = −6 Q(−3) = −9 + 3

Q(−3) = 3(−3) + 3 b. Q(−3)

R(2) = 2

2(2) − 4

R(2) = 2

4 − 4

R(2) = 2

0Undefined! 2 is not in the domain!

Example 1 d. P(−2) e. Q( 1

3 ) f. R(5)


3 ) f. R(5)

P(−2) = −8


3 ) f. R(5)

P(−2) = −8 Q( 13 ) = 4


3 ) f. R(5)

P(−2) = −8 Q( 13 ) = 4 R(5) = 5

6

Mapping Notation

Mapping NotationAlso known as Arrow Notation


A:x →


A:x →

Reads “A maps x onto”


A:x →

Reads “A maps x onto”

Still identifies the independent variable (after colon) and dependent variable (A:x)

Example 2Evaluate using the functions in Example 1.

a. P : 1 → b. Q : 2 → c. R : 3 →


a. P : 1 → b. Q : 2 → c. R : 3 →

P : 1 → 13


a. P : 1 → b. Q : 2 → c. R : 3 →

P : 1 → 13

= 1


a. P : 1 → b. Q : 2 → c. R : 3 →

P : 1 → 13

= 1

1


a. P : 1 → b. Q : 2 → c. R : 3 →

P : 1 → 13

= 1

1

Q : 2 → 3(2) + 3


a. P : 1 → b. Q : 2 → c. R : 3 →

P : 1 → 13

= 1

1

Q : 2 → 3(2) + 3

= 6 + 3


a. P : 1 → b. Q : 2 → c. R : 3 →

P : 1 → 13

= 1

1

Q : 2 → 3(2) + 3

= 6 + 3

= 9


a. P : 1 → b. Q : 2 → c. R : 3 →

P : 1 → 13

= 1

1

Q : 2 → 3(2) + 3

= 6 + 3

= 9

9


a. P : 1 → b. Q : 2 → c. R : 3 →

P : 1 → 13

= 1

1

Q : 2 → 3(2) + 3

= 6 + 3

= 9

9

R : 3 →

32(3) − 4


a. P : 1 → b. Q : 2 → c. R : 3 →

P : 1 → 13

= 1

1

Q : 2 → 3(2) + 3

= 6 + 3

= 9

9

R : 3 →

32(3) − 4

=

36 − 4


a. P : 1 → b. Q : 2 → c. R : 3 →

P : 1 → 13

= 1

1

Q : 2 → 3(2) + 3

= 6 + 3

= 9

9

R : 3 →

32(3) − 4

=

36 − 4

=

32


a. P : 1 → b. Q : 2 → c. R : 3 →

P : 1 → 13

= 1

1

Q : 2 → 3(2) + 3

= 6 + 3

= 9

9

R : 3 →

32(3) − 4

=

36 − 4

=

32

32

Example 3The area of a circle is a function of its radius.

Rewrite the formula using the following notations. A = πr2

a. Euler’s Notation b. Mapping Notation




A(r ) = πr2




A(r ) = πr2 A : r → πr2

Example 4If , then a : x → 2x + 7 a : 14 → ?


a : 14 → 2(14) + 7


a : 14 → 2(14) + 7 = 35


a : 14 → 2(14) + 7 = 35

a : 14 → 35

Example 5Suppose . f (x) = 4x2 − 2x + 9

Find . f (−3)


f (−3) = 4(−3)2 − 2(−3) + 9

Find . f (−3)


f (−3) = 4(−3)2 − 2(−3) + 9

= 4(9) + 6 + 9

Find . f (−3)


f (−3) = 4(−3)2 − 2(−3) + 9

= 4(9) + 6 + 9

= 36 + 6 + 9

Find . f (−3)


f (−3) = 4(−3)2 − 2(−3) + 9

= 4(9) + 6 + 9

= 36 + 6 + 9

= 51

Find . f (−3)


f (−3) = 4(−3)2 − 2(−3) + 9

= 4(9) + 6 + 9

= 36 + 6 + 9

= 51

f (−3) = 51

Find . f (−3)

Homework

Homework

p. 22 #1-26

AA Section 1-3

Health & Medicine

Transcript of AA Section 1-3