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Composite Functions Functions of Functions

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10/11/2013 Composite Functions 2 2 Fencing a Square Lot A square 2-acre yard is to be fenced How many feet of fencing is needed for each side? (1 acre = 43,560 sq.ft.) Solution: Converting acres to square feet: f(x) = 43560x sq.ft. f(2) = 43560(2) = sq.ft. = y sq.ft. Composition of Functions Let x = no. of acres y = no. of square ft.

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10/11/2013 Composite Functions 3 3 Fencing a Square Lot Solution: f(2) = 43560(2) = sq.ft. = y sq.ft. Converting sq.ft. to side length in feet: g(y) = y 1/2 ft. g(f(2)) = g(43560(2)) = (43560(2)) 1/2 = (87120) 1/2 = ft. Composition of Functions

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10/11/2013 Composite Functions 4 Fencing a Square Lot Solution: f(2) = y sq.ft. g(f(2)) = ft 4 y Square Feet x Acres Composition of Functions ● ● ● f g ? Feet g(f(x))

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10/11/2013 Composite Functions 5 Functions of Functions If y = f(x) and y is in the domain of g, then g(y) = g(f(x)) is called a composite function of x g is a function of a function 5 Composition of Functions

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10/11/2013 Composite Functions 6 Functions of Functions Definition For functions f and g, with g defined on the range of f, the function is the composite function of g composed with f 6 Composition of Functions (g f)(x) = g(f(x)) ° Note: g f ° is NOT the product of g and f

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10/11/2013 Composite Functions 7 Functions of Functions Domain of { x | x is in domain f and f(x) is in domain g } { x | x dom f ; f(x) dom g } Domain of 7 Composition of Functions is a subset of domain f (g f)(x) = g(f(x)) ° is (g f)(x) ° OR

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10/11/2013 Composite Functions 8 8 Stopping a Vehicle Find the reaction distance r(x) = tx traveled in estimated reaction time t = 2.5 secs at velocity x = 60 mph Converting mph to ft/hr: h(x) = 5280x ft/hr h(60) = 5280(60) = ft/hr = y ft/hr Composition of Functions

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10/11/2013 Composite Functions 9 9 Stopping a Vehicle h(60) = ft/hr = y ft/hr Converting ft/hr to ft/sec: g(y) = y/3600 ft/sec g(h(60)) = g(31600) = /3600 = 88 ft/sec = z ft/sec Composition of Functions

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10/11/2013 Composite Functions 10 Stopping a Vehicle h(60) = ft/hr = y ft/hr g(y) = g(h(60)) = 88 ft/sec = z ft/sec Converting ft/sec to feet: f(z) = (5/2)z ft Thus reaction distance at 60 mph is r(60) = f(g(h(60))) = f(g(316800)) = f(88) = (5/2)88 ft = 220 ft Composition of Functions

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10/11/2013 Composite Functions 11 Stopping a Vehicle h(60) = ft/hr = y ft/hr g(y) = g(h(60)) = 88 ft/sec = z ft/sec f(z) = (5/2)z ft = 220 ft r(60) = f(g(h(60))) = 220 ft Composition of Functions z ft/sec y ft/hr x mi/hr ● ● ● h g r(x) ● f 220 feet

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10/11/2013 Composite Functions 12 Composite Diagram: 12 Dom f Composition of Functions Range f f g Dom g Range g a f( a ) b g(b) Remember: Question: Is Dom g = Range f ?… not necessarily ! b Dom g but b Range f f(x) ≠ b for any x but g(b) Range g a Dom f but f( a ) Dom g (version 1) g f ° Range is a subset of Range g g f ° Dom is a subset of Dom f g f ° So ( )( a ) = g(f( a )) does not existg f ° x f(x) g(f(x)) ● ● ● ° Dom g f ° Range g f ° g f Where are Dom g f and Range g f ? ° °

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10/11/2013 Composite Functions 13 Domain f Composite Diagram Composition of Functions Range f ● ● ● x f(x) g(f(x)) f g Range g a f( a ) b g(b) (version 2) g f ° ° Dom g f Range g f ° Domain g Range is a subset of Range g g f ° Dom is a subset of Dom f g f ° °

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10/11/2013 Composite Functions 14 Example: 14 Composition of Functions f(x) = x + 2 g(x) = 5x + x = g(x + 2) = 5(x + 2) + x + 2 Dom f = R Dom g = { x x ≥ 0 } Range f = R Range g = { x x ≥ 0 } (g f)(x) = g(f(x)) °

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10/11/2013 Composite Functions 15 Example: 15 Composition of Functions f(x) = x + 2 g(x) = 5x + x = [ –2, ) = [ 0, ) = 5(x + 2) + x + 2 (g f)(x) ° Dom g f = { x x ≥ –2 } ° Range g f = { x x ≥ 0 } °

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10/11/2013 Composite Functions 16 Example: 16 Composition of Functions f(x) = x + 2 g(x) = 5x + x = f ( 5x + x ) = 5x + x + 2 (f g)(x) = f(g(x)) ° Question: (g f)(x) ° = 5(x + 2) + x + 2 ° How do g f and differ ? f g °

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10/11/2013 Composite Functions 17 Example: 17 Composition of Functions f(x) = x + 2 g(x) = 5x + x = 5x + x + 2 (f g)(x) ° = [ 2, ) Range f g ° (g f)(x) ° = 5(x + 2) + x + 2 Dom f g ° = [ 0, ) = [ –2, ) Dom g f ° = [ 0, ) Range g f ° What if f(x) = x 2 – 4 x – 2 = x + 2, for x ≠ 2 ?

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10/11/2013 Composite Functions 18 Example: 18 Composition of Functions f(x) = x + 2 g(x) = 5x + x f(x) = x 2 – 4 x – 2 = x + 2, for x ≠ 2 ? Dom f g ° = { x | x ≥ 0, x ≠.7350 } = { x | x ≥ –2, x ≠ 2 } Dom g f ° = { x | x ≥ 0, x ≠ 22 } Range g f ° Range f g ° = { x | x ≥ 2, x ≠ }

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10/11/2013 Composite Functions 19 Domain f Example: Composite Diagram Composition of Functions g(x) = 5x + x = g(x + 2) = 5(x + 2) + x + 2 Range f ● ● ● x f(x) g(f(x)) f g Domain g Range g –5 – f(x) = x 2 – 4 x – 2 = x + 2, for x ≠ 2, f(x) ≠ 4 f (g f)(x) = g(f(x)) ° g f ° Dom g f = { x x ≥ –2, x ≠ 2 } ° Range g f = { x x ≥ 0, x ≠ 22 } °

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10/11/2013 Composite Functions 20 Composition Versus Multiplication NOTE: Example : Let f(x) = x 2 and g(x) = x 1/2 Composition of Functions = g(x 2 ) = (x 2 ) 1/2 = |x|... then (gf)(x) = g(x) f(x) = x 1/2 x 2 = x 5/2... which is NOT |x| !!, for all x (g f)(x) ≠ (gf)(x) ° But … (g f)(x) = g(f(x)) ° But …

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10/11/2013 Composite Functions 21 Composition Versus Multiplication Is it possible that Composition of Functions... well... = f(g(x)) = f(x 1/2 ) = (x 1/2 ) 2 Question: Only if their domains are equal !Are they ?, for x ≥ 0 ? (g f)(x) = (f g)(x) ° ° Are and the same function ? g f ° f g ° = (g f)(x) ° (f g)(x) ° = x

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10/11/2013 Composite Functions 22 Find 22 Combining Functions Note: Output of the first function is input to the second = f(1) = 3 = -2= g(-1) andgraphically (g f)(–3) ° (f g)(2) ° = g(f(-3))(g f)(-3) ° = f(g(2))(f g)(2) ° f(1) = x y ● ● ● ● g(2) = 1 y = f(x) y =g(x) x y y = f(x) y =g(x) ● ● ● ● f(-3) = -1

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10/11/2013 Composite Functions 23 Combining Functions Find = f(g(2)) = g(f(-3)) Note: Output of the first function is input to the second = f(1) = 3 = -2= g(-1) andin tabular form (g f)(–3) ° (f g)(2) ° x –3 –2 – f(x) – g(x) 5 1 –2 –3 –2 1 5 –2 –1 3 1 –3 2 1 ° (f g)(2) ° (g f)(-3)

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10/11/2013 Composite Functions 24 Think about it !

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