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

P 279 DEF:  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) – 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 y 2a-5 3(2a–5) + 2 6a-13 a+b 3(a + b) + 2 3a+3b+2 a+h 3(a + h) + 2 3a+3h+2 a 3 a a + 2 3a+3h+2–(3a+2) = 3h 3a+3h+2–3a–2= 3h

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

EX: If g ( x ) =√ x - 1; h ( x ) = x 2 +1 g ( x ) =√x–1 g (1) = √1– 1 √ 0 0 _____________ g(1) – h(3) = h ( x ) = x 2 +1 h (3) = _____________ 0 – 10 = -10

EX: If g ( x ) =√ x - 1; h ( x ) = x 2 +1 √x – 1 g (5) √5 – 1 √ 4 2 _____________ h (2) + g(5) = h ( x ) = x 2 +1 h (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 of g ( x ) = x x + 6 = (x – 2) (x – 3) x 2 – 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 x 2 – 3 FIND: f ( g ( - 4 ) ) f º g(- 4) g(- 4) = 2(- 4) 2 – 3 = 2 · 16 – 3 = 32 – 3 = 29 f(29) = = 31 HW 5:P281(27-30,45-48) FIND: g ( f ( - 4 )) g º f (-4) f (-4) = = -2 g (-2) = 2(-2) 2 – 3 = 8 – 3 = 5

EX: If g ( x ) =  x – 1 ; h ( x ) = x FIND: h ( g ( 2 ) ) h º g(2) g(2) =  2 – 1= 1 h(1) = = 2 FIND: g ( h ( 3 ) ) g º h(3) h(3) = = 10 g(10)=  10 – 1 = 3