lecture19(complete)

lecture19(completeï&frac...

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MA1100 Lecture 19 Functions omposition of functions Composition of functions Inverse functions Chartrand: 9.5, 9.6
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P l ( L k i f r r d ) Lesson Plan (Looking forward) Tue Lecture Fri Lecture Tutorial Week 11 20. Number Th. HW4 21. Number Th. Tut 8 Functions Week 12 22. Number Th. 23. Cardinality Tut 9 Number Th. Week 13 24. Revision 25. Revision Tut 10 lecture HW5 lecture (tentative) Number Th./ Cardinality Week 14 EADING WEEK READING WEEK Week 15 FINAL EXAM (NOV 24)
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p i t i f f t i Composition of functions Example f: R Ø R defined by f(x) = x 2 g: R Ø R defined by g(x) = x + 1 he omposition f f and g The resulting function has formula 2 1 The composition of f and g a formula x + 1 Lecture 19 3
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p i t i f f t i Composition of functions Example f: R Ø R defined by f(x) = 3x 2 + 2 g: R Ø R defined by g(x) = sin(x) R R f f(x) = 3x 2 + 2 g( ) g( ) = sin( 3x 2 + 2 ) R g 3x 2 + 2 x sin(3x 2 + 2) g(x) = sin(x) f( ) f( ) = 3 sin(x) 2 + 2 R R g R f Lecture 19 4 sin(x) x 3sin 2 x + 2
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p i t i f f t i Composition of functions efinition Definition f: A Ø B and g: B Ø C be functions. he omposition f f and g The composition of f and g is the function g o f : A Ø C defined by (g o f)(x) = g(f(x)) for all x œ A. We say g o f is a composite function . inner function outer function f(x) x A B f g(f(x)) C g Lecture 19 5 g o f
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p i t i f f t i Composition of functions emarks Remarks ay not be defined A f: A Ø B and g: B Ø C be functions. • f o g may not be defined if A C g o f is not equal to f o g •“ omposition of f and g ” not the same as co pos t o o a d g ot t e sa e as composition of g and f g f f(x) x A B f g(f(x)) C g g(x) x B C f(g(x))? B g(x)? A Lecture 19 6 g o f f o g ?
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i t i L Associative Law xample (3 or more functions) Chartrand: Theorem 9.9 Example (3 or more functions) f: A Ø B, g: B Ø C, h: C Ø D A f: R Ø R f(x) = x 2 g: R Ø R in(x g o f : A Ø C h o ( g o f ): A Ø D g(x) = sin x h: R Ø R sin(x 2 ) f: A Ø B, g: B Ø C, h: C Ø D h(x) = x 1/3 = in(x /3 h o g : B Ø D ( h o g ) o f : A Ø D h o ( g o f ) = [ sin(x 2 ) ] 1/3 ( h o g ) o f = [sin( x 2 )] 1/3 [sin(x)] 1/3 Lecture 19 7 Associative law: h o (g o f) = (h o g) o f
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p i f t i Decomposing functions xample Example e can ecompose as a composite function g F: R Ø R defined by F(x) = x 2 + 1 . We can decompose F as a composite function g o f f: R Ø R defined by f(x) = x 2 g: R Ø R defined by g(x) = x + 1 Decompose K in terms of f and g: K: R Ø R defined by K(x) = (x 2 + 1) 2 .
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This note was uploaded on 04/15/2010 for the course MATHS MA1101R taught by Professor Vt during the Spring '10 term at National University of Singapore.

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lecture19(completeï&frac...

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