11.2 Values and Composition of Functions

OBJ: Find the value of a function, given an element in the domain of the function

Find the range and domain of a function

? What is the function graphed below?

What is the domain?

What is the range?

(Top p279; Rev. ex.)

{(– 3, -2 ), (–2, 3), (3, 2), (4, 1)}

{– 3, – 2, 3, 4}

{– 2, 1, 2, 3}

y

x

5

5

-5-5

P 279DEF: Value of a function:f ( x ) = y

READ: “f at (or “f of ) x equals y”

EX1 g = {(-1,-2), (0,3), (2,3), (3,-1)}

FIND: g (-1)= g (0)=

g (2)= g (3)=

HW4: p281: (2-6e)

g (-1)

= -2

g (0)

= 3

g (2)

= 3

g (3)

= -1

NOTE: A function “f” is often defined by giving an equation or formula for its range.P279

EX 2 If f (x) = 2 x–3, find f (1), f (2), f (3)

HW 4: P281: (8-16e)

x 2x– 3 f(x) or y

1

2(1) – 3

– 1

2

2(2) – 3

1

3

2(3) – 3

3

EX3 If f ( x ) = 3 x + 2 P280 HW 5: P281(18, 37-44, 49-52)FIND: f ( 2 a – 5 ) f ( a + b ) f ( a + h ) - f ( a )

x 3x +2 f(x) or y2a-5 3(2a–5) + 2 6a-13a+b 3(a + b) + 2 3a+3b+2a+h 3(a + h) + 2 3a+3h+2 a 3 a + 2 3 a + 2 3a+3h+2–(3a+2) = 3a+3h+2–3a–2= 33hh

P 280 HW4: P281: (20-26e)EX:4 • If h ( x ) = – x2 + 3, find the range of h for the domain D = {– 2 , 0 , 1 } x –x2 + 3 h(x)

–2 – (-2 )2 + 3 – 1

0 – ( 0 )2 + 3 3

1 – ( 1 )2 + 3 2Range R = {-1. 2, 3}

EX: • If g ( x ) =√ x - 1; h ( x ) = x2+1

g ( x ) =√x–1 g (1) =

√1– 1

√ 0

0

_____________

g(1) – h(3) =

h ( x ) = x2 +1 h (3) =

32 + 1

9 + 1

10

_____________

0 –

10 = -10

EX: • If g ( x ) =√ x - 1; h ( x ) = x2+1

√x – 1g (5) √5 – 1 √ 4 2_____________ h (2) + g(5) =

h ( x ) = x2 +1 h (2) 22 + 1 4 + 1 5 _____________ 5 + 2 = 7

11.2 Composition of Functions

OBJ: To find a value of a function that is composed of two other functions

DEF: Rational Function Quotient of two polynomial functions

NOTE: Domain is the set of all real numbers x Ʀ, except those numbers that make the denominator equal to zero.

P 280 HW 5: P 281: (31-36)Ex 5? What is the domain ofg ( x ) = x2 - 5 x + 6 = (x – 2) (x – 3) x2 – 2 x x(x – 2) {x | x Ʀ, x≠0, 2} or {all real numbers except 0 or 2} (-∞, 0) U (0, 2) U (2, ∞)

EX: 7 If f ( x ) = x + 2;

g ( x ) = 2 x2 – 3

FIND: f ( g ( - 4 ) )

f º g(- 4)

g(- 4) =

2(- 4)2 – 3 =

2 · 16 – 3 =

32 – 3 = 29

f(29) =

29 + 2 = 31

HW 5:P281(27-30,45-48)

FIND: g ( f ( - 4 ))

g º f (-4)

f (-4) =

-4 + 2 = -2

g (-2) =

2(-2)2 – 3 =

8 – 3 = 5

EX: If g ( x ) = x – 1 ; h ( x ) = x2 + 1FIND: h ( g ( 2 ) )

h º g(2)

g(2) =

2 – 1=

1

h(1) =

12 +

1=

2

FIND: g ( h ( 3 ) )

g º h(3)

h(3) =

32 +

1=

10

g(10)=

10 –

1 =

3