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### Concept map function

• 1. TOPIC : FUNCTIONS Absolute Value Functions Eg: f (x) = x − 2 Function Notations 1) Find the possible values of x • f :x → x+2 if f(x) = 3 CONCEPTS MAP (means the function f maps x onto ” x + 2 Solution x-2 =3 , x=5 • f ( x) = x + 2 is read as ’ f of x is x + 2 - (x - 2) = 3 , x = -1 RELATIONS • x is the object , f (x) is the image 2) Sketch the Graph of Examples: a) If 5 is the object , find the image f (x) = x − 2 for the Solution : f ( 5) = 5 + 2 = 7, domain -1 ≤ x ≤ 6 How to represent Types Of Relations Solution: Relations b) Given f (3 y) = 11 , find y. x = -1 , f(x) = 3 1. One-to-one relation Solution: f ( 3y) = 3y + 2 = 11, x = 5 , f(x) = 3 1) Arrow diagram 3y=9 , f(x) = x-3 = 0, x = 3 y=3 f (x) a• •1 2 4 3 5 F 3 b• 6 7 U Composite Functions •2 N • If a function f is followed by a function g , x c• •3 we obtain the composite function g f . -1 0 3 6 2. One-to-many relation C 3) Corresponding f g d• •4 T • Range: x f (x) gf 0 ≤ f(x) ≤ 3 2) Ordered Pairs 1 4 I gf (a,1) (b,2), (c,2), 3 5 O Inverse functions (f-1) (d,3) 6 • In general gf ≠ fg . N • Concept: f(x) = y , Then, f –1 (y) = x 3 3) Graph S • How to determine composite function: • Eg: 3. Many-to-one relation Example : Given f : x  x +1 Given f : x  2x + 1. Find f –1 2 × g : x  2x Solution: 1 × × 7 6 Determine i.) f g ii ) f 2 y −1 y = 2x + 1 , x = s 9 Solutions; 2 × 10 i.) fg (x) = f (g (x) (ii) f 2 (x) = f f (x) x −1 11 14 = f ( 2x ) = f (x +1) So ; f –1 (x) = a b c d 2 = 2x + 1 = (x +1) +1 = x + 2 •Note: Only one – to – one functions will Objects : a, b, c, d Eg.• Given f : x  2x and give one – to – one inverse functions. 4. Many-to-many relation Domain : {a, b, c, d} g f: x  3x - 1, find g. • ff-1(x) = f -1f (x) = x Solution: g (f(x) = 3 x -1 Eg.• Given f : x  2x and Codomain : {1, 2, 3,4} 1 5 g( 2x) = 3x - 1 f g: x  x + 3, find g. f(x) = 2x+1 x −1 Solution: f (g(x) ) = x + 3 f −1( x ) = y 2 Images : 1,2,3 7 Lets 2x = y, x= , 2 g(x) = x + 3 2 2 ax + b − dx + b Range : { 1, 2, 3 } x+3 f ( x) = f ( x) = y 3x g ( x) = cx + d cx − a g(y) = 3( ) -1,g(x) = −1 2 2 2
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