x 2 if 2 x < 0 4 x if 2 x 6

(1)

f(x) =                   

x2 if − 2 ≤ x < 0

2x if 0 ≤ x < 2

(2)

f(x) =                   

x2 if − 2 ≤ x < 0

2x if 0 ≤ x < 2

4 − x if 2 ≤ x ≤ 6

Find the value of

(1) f(−1) (2) f 1

2 (3) f(2.5)

(3)

f(x) =                   

x2 if − 2 ≤ x < 0

2x if 0 ≤ x < 2

4 − x if 2 ≤ x ≤ 6

Find the value of

(1) f(−1) = (−1)2 = 1 (2) f 1

2 (3) f(2.5)

(4)

f(x) =                   

x2 if − 2 ≤ x < 0

2x if 0 ≤ x < 2

4 − x if 2 ≤ x ≤ 6

Find the value of

(1) f(−1) = (−1)2 = 1 (2) f 1

2

= 2 · 1

2 = 1

(3) f(2.5)

(5)

f(x) =                   

x2 if − 2 ≤ x < 0

2x if 0 ≤ x < 2

4 − x if 2 ≤ x ≤ 6

Find the value of

(1) f(−1) = (−1)2 = 1 (2) f 1

2

= 2 · 1

2 = 1

(3) f(2.5) = 4 − 2.5 = 1.5 and sketch the graph of f.

(6)

f(x) =                   

x2 if − 2 ≤ x < 0

2x if 0 ≤ x < 2

4 − x if 2 ≤ x ≤ 6

Find the value of

(1) f(−1) = (−1)2 = 1 (2) f 1

2

= 2 · 1

2 = 1

Solution To draw the graph of f, divide the domain [−2,6] into three subintervals: [−2,0), [0,2) and [2,6].

(7)

f(x) =                   

x2 if − 2 ≤ x < 0

2x if 0 ≤ x < 2

4 − x if 2 ≤ x ≤ 6

Find the value of

(1) f(−1) = (−1)2 = 1 (2) f 1

2

= 2 · 1

2 = 1

-2 2 4 6

-2 -1 1 2 3 4

(8)

f(x) =                   

x2 if − 2 ≤ x < 0

2x if 0 ≤ x < 2

4 − x if 2 ≤ x ≤ 6

Find the value of

(1) f(−1) = (−1)2 = 1 (2) f 1

2

= 2 · 1

2 = 1

-2 2 4 6

-2 -1 1 2 3 4

(9)

f(x) =                   

x2 if − 2 ≤ x < 0

2x if 0 ≤ x < 2

4 − x if 2 ≤ x ≤ 6

Find the value of

(1) f(−1) = (−1)2 = 1 (2) f 1

2

= 2 · 1

2 = 1

-2 2 4 6

-2 -1 1 2 3 4

(10)

|x| =        

x if x ≥ 0

(11)

|x| =        

x if x ≥ 0

−x if x < 0 Example

(12)

|x| =        

x if x ≥ 0

(1) |2| = 2

(13)

|x| =        

x if x ≥ 0

(1) |2| = 2

(14)

|x| =        

x if x ≥ 0

(1) |2| = 2

(2) |−3| = −(−3) = 3

(15)

|x| =        

x if x ≥ 0

(1) |2| = 2

(2) |−3| = −(−3) = 3

• |a| is the distance from a to 0.

> | −3 | 0 | 2 3units

(16)

|x| =        

x if x ≥ 0

(1) |2| = 2

(2) |−3| = −(−3) = 3

• |a| is the distance from a to 0.

> | −3 | 0 | 2 3units

z }| {z }|2units {

(17)

(18)

> |

−a

| 0

| a z }| {z }| {

(19)

> |

−a

| 0

| a z }| {z }| {

(20)

> |

−a

| 0

| a z }| {z }| {

• |a − b| is the distance from a to b. Example a = −3, b = 2

(21)

> |

−a

| 0

| a z }| {z }| {

(22)

> |

−a

| 0

| a z }| {z }| {

(23)

> |

−a

| 0

| a z }| {z }| {

|−3 − 2| = | − 5|

(24)

> |

−a

| 0

| a z }| {z }| {

|−3 − 2| = | − 5|

(25)

> |

−a

| 0

| a z }| {z }| {

|−3 − 2| = | − 5|

= 5 = distance from −3 to 2

(26)

> |

−a

| 0

| a z }| {z }| {

|−3 − 2| = | − 5|

• √a2 ₌ _|a|

(27)

> |

−a

| 0

| a z }| {z }| {

|−3 − 2| = | − 5|

• √a2 ₌ _|a|

(28)

> |

−a

| 0

| a z }| {z }| {

|−3 − 2| = | − 5|

• √a2 ₌ _|a|

Example √32 ₌ √₉ ₌ ₃ ₌ _|₃_| p

(29)

> |

−a

| 0

| a z }| {z }| {

|−3 − 2| = | − 5|

• √a2 ₌ _|a|

Example √32 ₌ √₉ ₌ ₃ ₌ _|₃_| p

(30)

(31)

Graph of the absolute value function f(x) = |x|

Divide the real number line into two subintervals [0,∞) and (−∞,0)

f(x) =         

x if x ≥ 0 −x if x < 0

(32)

f(x) =         

x if x ≥ 0 −x if x < 0

-2 -1 1 2

0.5 1 1.5

(33)

f(x) =         

x if x ≥ 0 −x if x < 0

-2 -1 1 2

0.5 1 1.5

(34)

(35)

y =

          

1 − x if x ≥ 0

(36)

y =

          

1 − x if x ≥ 0

1 − (−x) if x < 0

-2 -1 1 2

-1 -0.5

0.5 1

(37)

y =

          

1 − x if x ≥ 0

1 − (−x) if x < 0

-2 -1 1 2

-1 -0.5

0.5 1

(38)

y =

          

1 − x if x ≥ 0

1 − (−x) if x < 0

-2 -1 1 2

-1 -0.5

0.5 1

(39)

y =

          

1 − x if x ≥ 0

1 − (−x) if x < 0

-2 -1 1 2

-1 -0.5

0.5 1

(40)

y =

          

1 − x if x ≥ 0

1 − (−x) if x < 0

-2 -1 1 2

-1 -0.5

0.5 1

(41)

(42)

Example Graph y = |x − 1|

y =

          

x − 1 if x − 1 ≥ 0 −(x − 1) if x − 1 < 0

(43)

y =

          

x − 1 if x − 1 ≥ 0 −(x − 1) if x − 1 < 0

-1 1 2 3

0.5 1 1.5

(44)

y =

          

x − 1 if x − 1 ≥ 0 −(x − 1) if x − 1 < 0

-1 1 2 3

0.5 1 1.5

(45)

Example Graph y = |x − 1| y =           

x − 1 if x − 1 ≥ 0 −(x − 1) if x − 1 < 0

-1 1 2 3

0.5 1 1.5

2

-2 -1 1 2 3

0.5 1 1.5

(46)

x − 1 if x − 1 ≥ 0 −(x − 1) if x − 1 < 0

-1 1 2 3

0.5 1 1.5

2

-2 -1 1 2 3

0.5 1 1.5

(47)

x − 1 if x − 1 ≥ 0 −(x − 1) if x − 1 < 0

-1 1 2 3

0.5 1 1.5

2

-2 -1 1 2 3

0.5 1 1.5

(48)

(49)

(50)

• First, square both sides to get y2 = x.

-2 -1 1 2 3 4

-2 -1

1 2 3

4 y = x 2

(51)

• First, square both sides to get y2 = x. Its graph is a parabola.

-2 -1 1 2 3 4

-2 -1

1 2 3

4 y = x 2

(52)

• Squarring introduces extra points.

-2 -1 1 2 3 4

-2 -1

1 2 3

4 y = x 2

(53)

√

x is always non-negative.

-2 -1 1 2 3 4

-2 -1

1 2 3

4 y = x 2

(54)

√

Graph of y = √x is the upper half of the parabola.

-2 -1 1 2 3 4

-2 -1

1 2 3

4 y = x 2

x = y2

(55)

√

-2 -1 1 2 3 4

-2 -1

1 2 3

4 y = x 2

x = y2

y = √x

Note y = √x

(56)

√

-2 -1 1 2 3 4

-2 -1

1 2 3

4 y = x 2

x = y2

y = √x

Note y = √x

y2 = x, y ≥ 0 Reason a = b =⇒ a2 = b2

(57)

√

-2 -1 1 2 3 4

-2 -1

1 2 3

4 y = x 2

x = y2

y = √x

Note y = √x

y2 = x, y ≥ 0 Reason a = b =⇒ a2 = b2

(58)

(1) y = √x − 2 Solution

(59)

(60)

The graph is obtained by moving that of y = √x two units down.

2.5 5 7.5 10 12.5 15

-2 -1

1 2 3

4 y =

√

(61)

2.5 5 7.5 10 12.5 15

-2 -1

1 2 3

4 y =

√

(62)

2.5 5 7.5 10 12.5 15

-2 -1

1 2 3

4 y =

√

x

(63)

(64)

(65)

The graph is obtained by moving that of y = √x two units to the right.

2 4 9 16

1 2 3 4

(66)

2 4 9 16

1 2 3 4

(67)

2 4 9 16

1 2 3 4

y = √x

(68)

2 4 9 16

1 2 3 4

y = √x

y = √x − 2

√

(69)

(70)

Eg. we may combine the square function and sine function to get sin(x2)

(71)

Eg. we may combine the square function and sine function to get sin(x2) x 7→ x2 7→ sin(x2)

(72)

Definition Let f and g be functions. The composition of g with f, denoted by g ◦ f, is the function defined by

(73)

(g ◦ f)(x) = g f(x)

(74)

(g ◦ f)(x) = g f(x) Example Let f and g be functions given by

f(x) = x2 g(x) = sin x

Then g ◦ f : R −→ R is

(75)

Then g ◦ f : R −→ R is (g ◦ f)(x) = g(x2)

(76)

Then g ◦ f : R −→ R is (g ◦ f)(x) = g(x2)

= sin(x2)

(77)

Example Let f(x) = √x and g(x) = x − 2. Find the following (if defined)

(1) (g ◦ f)(9)

(2) (f ◦ g)(6)

(78)

(1) (g ◦ f)(9) = g f(9)

(2) (f ◦ g)(6)

(79)

(1) (g ◦ f)(9) = g f(9)

= g(3)

(2) (f ◦ g)(6)

(80)

(1) (g ◦ f)(9) = g f(9)

= g(3) = 1 (2) (f ◦ g)(6)

(81)

(1) (g ◦ f)(9) = g f(9)

= g(3) = 1

(2) (f ◦ g)(6) = f g(6)

(82)

(1) (g ◦ f)(9) = g f(9)

= g(3) = 1

(2) (f ◦ g)(6) = f g(6)

= f(4)

(83)

(1) (g ◦ f)(9) = g f(9)

= g(3) = 1

(2) (f ◦ g)(6) = f g(6)

= f(4) = 2 (3) (f ◦ g)(1)

(84)

(1) (g ◦ f)(9) = g f(9)

= g(3) = 1

(2) (f ◦ g)(6) = f g(6)

= f(4) = 2

(85)

(1) (g ◦ f)(9) = g f(9)

= g(3) = 1

(2) (f ◦ g)(6) = f g(6)

= f(4) = 2

(3) (f ◦ g)(1) = f g(1)

(86)

(1) (g ◦ f)(9) = g f(9)

= g(3) = 1

(2) (f ◦ g)(6) = f g(6)

= f(4) = 2

(3) (f ◦ g)(1) = f g(1)

(87)

(1) (g ◦ f)(9) = g f(9)

= g(3) = 1

(2) (f ◦ g)(6) = f g(6)

= f(4) = 2

(3) (f ◦ g)(1) = f g(1)

= f(−1) undefined

(88)

Example Let f(x) = x2 and g(x) = 2x + 1. Find

(1) (f ◦ g)(x) (2) (g ◦ f)(x)

(89)

(1) (f ◦ g)(x) (2) (g ◦ f)(x) Solution

(90)

(1) (f ◦ g)(x) (2) (g ◦ f)(x) Solution

(1) (f ◦ g)(x) = f(g(x)) = f(2x + 1)

(91)

(1) (f ◦ g)(x) (2) (g ◦ f)(x) Solution

(1) (f ◦ g)(x) = f(g(x)) = f(2x + 1) = (2x + 1)2

(92)

(1) (f ◦ g)(x) (2) (g ◦ f)(x) Solution

(1) (f ◦ g)(x) = f(g(x)) = f(2x + 1) = (2x + 1)2 = 4x2 + 4x + 1

(93)

(1) (f ◦ g)(x) (2) (g ◦ f)(x) Solution

(1) (f ◦ g)(x) = f(g(x)) = f(2x + 1) = (2x + 1)2 = 4x2 + 4x + 1

(94)

(1) (f ◦ g)(x) (2) (g ◦ f)(x) Solution

(1) (f ◦ g)(x) = f(g(x)) = f(2x + 1) = (2x + 1)2 = 4x2 + 4x + 1

(2) (g ◦ f)(x) = g(f(x)) = g(x2)

(95)

(1) (f ◦ g)(x) (2) (g ◦ f)(x) Solution

(1) (f ◦ g)(x) = f(g(x)) = f(2x + 1) = (2x + 1)2 = 4x2 + 4x + 1

(2) (g ◦ f)(x) = g(f(x)) = g(x2) = 2x2 + 1

(96)

(1) (f ◦ g)(x) (2) (g ◦ f)(x) Solution

(1) (f ◦ g)(x) = f(g(x)) = f(2x + 1) = (2x + 1)2 = 4x2 + 4x + 1

(2) (g ◦ f)(x) = g(f(x)) = g(x2) = 2x2 + 1

(97)

f(x) = 2x + 1, g(x) = x − 1 2 . Find

(1) (f ◦ g)(x) (2) (g ◦ f)(x)

(98)

f(x) = 2x + 1, g(x) = x − 1 2 . Find

(1) (f ◦ g)(x) = x (2) (g ◦ f)(x) = x

(99)

f(x) = 2x + 1, g(x) = x − 1 2 . Find

(1) (f ◦ g)(x) = x (2) (g ◦ f)(x) = x

Definition A function ϕ from a set X into itself satisfying ϕ(x) = x for all x ∈ X is called the identity function on X.

(100)

f(x) = 2x + 1, g(x) = x − 1 2 . Find

(1) (f ◦ g)(x) = x (2) (g ◦ f)(x) = x

Remark The above example means that (f ◦ g) and (g ◦ f) are the identity functions on R.

(101)

f(x) = 2x + 1, g(x) = x − 1 2 . Find

(1) (f ◦ g)(x) = x (2) (g ◦ f)(x) = x

Remark The above example means that (f ◦ g) and (g ◦ f) are the identity functions on R.

Question Given a function f : X −→ Y, when can we find a function g : Y −→ X

(102)

1 2 3 4 5 6 7 A

B C D E

(103)

1 2 3 4 5 6 7 A

B C D E

Is there any g : Y −→ X satisfying (g ◦ f)(x) = x for all x ∈ X ?

(104)

1 2 3 4 5 6 7 A

B C D E

(105)

1 2 3 4 5 6 7 A

B C D E

Solution Yes. Define g(1) = A g(2) = C

(106)

1 2 3 4 5 6 7 A

B C D E

Solution Yes. Define g(1) = A g(2) = C

Then (g ◦ f)(A) = A

(107)

1 2 3 4 5 6 7 A

B C D E

Solution Yes. Define g(1) = A g(2) = C g(3) = D

(108)

1 2 3 4 5 6 7 A

B C D E

Solution Yes. Define g(1) = A g(2) = C g(3) = D

(g ◦ f)(C) = C (g ◦ f)(D) = D

(109)

1 2 3 4 5 6 7 A

B C D E

Solution Yes. Define g(1) = A g(2) = C g(3) = D g(4) = A

(g ◦ f)(C) = C (g ◦ f)(D) = D

(110)

1 2 3 4 5 6 7 A B C D E

Solution Yes. Define g(1) = A g(2) = C g(3) = D g(4) = A g(5) = B

(g ◦ f)(C) = C (g ◦ f)(D) = D

(111)

1 2 3 4 5 6 7 A B C D E

Solution Yes. Define g(1) = A g(2) = C g(3) = D g(4) = A g(5) = B

Then (g ◦ f)(A) = A (g ◦ f)(B) = B (g ◦ f)(C) = C (g ◦ f)(D) = D

(112)

1 2 3 4 5 6 7 A B C D E

Solution Yes. Define g(1) = A g(2) = C g(3) = D g(4) = A g(5) = B g(6) = A

(113)

1 2 3 4 5 6 7 A B C D E

Solution Yes. Define g(1) = A g(2) = C g(3) = D g(4) = A g(5) = B g(6) = A g(7) = E

(114)

1 2 3 4 5 6 7 A B C D E

Solution Yes. Define g(1) = A g(2) = C g(3) = D g(4) = A g(5) = B g(6) = A g(7) = E

Then (g ◦ f)(A) = A (g ◦ f)(B) = B (g ◦ f)(C) = C (g ◦ f)(D) = D (g ◦ f)(E) = E

(115)

1 2 3 4 5 6 7 A

B C D E

(116)

1 2 3 4 5 6 7 A

B C D E

(117)

1 2 3 4 5 6 7 A

B C D E

Solution Need (g ◦ f)(A) = A

(118)

1 2 3 4 5 6 7 A

B C D E

(g ◦ f)(B) = B (g ◦ f)(C) = C

(119)

1 2 3 4 5 6 7 A

B C D E

(g ◦ f)(B) = B (g ◦ f)(C) = C

(120)

1 2 3 4 5 6 7 A

B C D E

(g ◦ f)(B) = B (g ◦ f)(C) = C

(g ◦ f)(D) = D (g ◦ f)(E) = E

(121)

1 2 3 4 5 6 7 A

B C D E

(g ◦ f)(B) = B (g ◦ f)(C) = C

(g ◦ f)(D) = D (g ◦ f)(E) = E

(122)

1 2 3 4 5 6 7 A

B C D E

(g ◦ f)(B) = B (g ◦ f)(C) = C

(g ◦ f)(D) = D (g ◦ f)(E) = E

In particular, g(1) = A g(1) = C

(123)

1 2 3 4 5 6 7 A B C D E

(g ◦ f)(B) = B (g ◦ f)(C) = C

(g ◦ f)(D) = D (g ◦ f)(E) = E

In particular, g(1) = A g(1) = C

(124)

(125)

x1 , x2 =⇒ f(x1) , f(x2) Example

(126)

x1 , x2 =⇒ f(x1) , f(x2) Example

f(x) = 2x is injective

-2 -1 1 2

1 2 3 4

(127)

x1 , x2 =⇒ f(x1) , f(x2) Example

-2 -1 1 2

1 2 3 4

(128)

x1 , x2 =⇒ f(x1) , f(x2) Example

-2 -1 1 2

1 2 3 4

x₁ , x₂ =⇒ 2x1 , 2x2

(129)

x1 , x2 =⇒ f(x1) , f(x2) Example

-2 -1 1 2

1 2 3 4

x₁ , x₂ =⇒ 2x1 , 2x2

g(x) = x2 is NOT injective

-2 -1 1 2

1 2 3 4

(130)

x1 , x2 =⇒ f(x1) , f(x2) Example

-2 -1 1 2

1 2 3 4

x₁ , x₂ =⇒ 2x1 , 2x2

-2 -1 1 2

1 2 3 4

(131)

x1 , x2 =⇒ f(x1) , f(x2) Example

-2 -1 1 2

1 2 3 4

x₁ , x₂ =⇒ 2x1 , 2x2

-2 -1 1 2

1 2 3 4

−1 , 1 but (−1)2 = 12

Remark Geometrically, f is injective means that graph of f intersects every horizontal line in at most one point.