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**Funcitions of Functions Functions of Functions**

Composite Functions Functions of Functions Combining Functions Once simple functions are defined, more complex functions can be constructed from them by combing then in various ways. The first set of combinations are simply the arithmetic operations that we normally use on numbers. The next area we look into is the composition of functions, that is, considering a function of another function. This leads eventually to discussion of inverse functions. Composite Functions 10/11/2013

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**Composition of Functions**

Funcitions of Functions 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. Let x = no. of acres y = no. of square ft. Fencing a Square Lot We note that there are several conversions possible in this kind of problem, depending on the units used. For example, 1 acre = 160 sq. rods = 4840 sq. yds. = 43,560 sq. ft. where 1 rod = 16.5 ft. = 5.5 yds. Thus we could have computed Area = 2 acres = 2(160 sq. rods) = 320 sq. rods, side = rods. Or we could have computed area and side as Area = 2 acres = 2(4840 sq. yds.) = 9680 sq. yds. side = = yds. The point to make here is that the second conversion, in any of its forms, depends on the output of the first conversion. Hence, the combined conversion is in fact a function of a function. We define function f as the conversion function from acres to square feet. Thus f converts x acres to y square feet, or just f(x) = y. Using the conversion factor of 1 acre = 43,560 sq.ft. we see that f(2) = 87,120 10/11/2013 Composite Functions 2 Composite Functions 10/11/2013

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**Composition of Functions**

Funcitions of Functions Fencing a Square Lot Solution: f(2) = 43560(2) = sq.ft. = y sq.ft. Converting sq.ft. to side length in feet: g(y) = y1/2 ft. g(f(2)) = g(43560(2)) = (43560(2))1/2 = (87120)1/2 = ft. Fencing a Square Lot We a second function g to convert square feet to linear perimeter feet. Thus we have y square feet converted to z linear feet, or just g(y) = z Using the relationship y = z2 we see that z = y½ and thus z = g(y) = y½ Putting this together with our previous result for y = f(x) we have z = g(y) = g(f(x)) or z = g(f(2)) = g(43560(2)) = (87120)½ = 295.1 The point to make here is that the second conversion g depends on the output of the first conversion that is g is a function of f. Hence, the combined conversion is in fact a function of a function. 3 10/11/2013 Composite Functions Composite Functions 10/11/2013

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**Composition of Functions**

Funcitions of Functions Fencing a Square Lot Solution: f(2) = y sq.ft. g(f(2)) = ft f g ● ● ● Fencing a Square Lot Here we illustrate the fencing problem with diagrams showing the domains and ranges of the functions already described. The first function, that is f, converts acres to square feet. The second function, g, converts square feet to linear feet on each side of the square perimeter. 2 87120 295.16 ? g(f(x)) x Acres y Square Feet Feet Composite Functions 4 10/11/2013 Composite Functions 10/11/2013

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**Composition of Functions**

Funcitions of Functions 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 Composition of Functions Composition is a special form of combining functions. Instead of f and g acting simultaneously on the same domain element x, we apply the functions in consecutive fashion, the first function acting on x and the second acting on the image of x under the first function. In the example, f acts on any x in its domain, then g acts on f(x), not on x itself. The caveat here is that f(x) must lie in the domain of g. We can then define the composite function acting on an element x in the domain of f and producing an element in the range of g. Thus, the domain of is a subset of the domain of f. We might also note that, since g carries the values of f(x) lying in the domain of g into the range of g, then the range of is a subset of the range of g. 5 10/11/2013 Composite Functions Composite Functions 10/11/2013

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**Composition of Functions**

Funcitions of Functions 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 (g f)(x) = g(f(x)) Functions of Functions Here we formally define what is meant by composite function. Up to now we have focused on how composite functions behave, that is, what these function do. We now focus on what a composite function is. We also introduce the special notation for composite functions, that is which emphasizes the function-of-a-function notion. Note: g f is NOT the product of g and f 10/11/2013 Composite Functions 6 Composite Functions 10/11/2013

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**Composition of Functions**

Funcitions of Functions 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 } (g f)(x) = g(f(x)) is OR Functions of Functions Since the function f has values f(x), we note that x must be in the domain of f and f(x) must be in the range of f. For the composite g(f(x)) to exist we see that f(x) must also lie in the domain of g. Following this line of thought, we see that g(f(x)) depends indirectly on x ; that is g(f(x)) is a function of x – the same x that gives f(x) its value. This makes the∩∞≠|≥≤•~½ domain of the composite function g(f(x)) a subset of the domain of f. Thus Dom (g f) = { x | x Dom f ; f(x) Dom g } Dom f (g f)(x) is a subset of domain f ∩ 10/11/2013 Composite Functions 7 Composite Functions 10/11/2013

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**Composition of Functions**

Funcitions of Functions 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 Stopping a Vehicle We note that for a vehicle traveling x miles per hour (mi/hr or mph), stopping distance s(x) is composed of two components: reaction distance r(x), the distance traveled while the driver reacts to a need to stop and hits the brake, and the braking distance b(x), the distance traveled while the brakes are actually applied. Test trials indicate the average human reaction time is about 2.5 seconds. As part of the overall stopping distance calculation, we want to compute the reaction distance, expressed as a function of the speed of the vehicle given in miles per hour. Since we are dealing with reaction times in seconds and want to know how many feet the vehicle travels (not how many miles), we must apply conversion factors to convert miles to feet and hours to seconds. The conversion factors simplify to the constant 11/3 which converts velocity x in mph to feet traveled in the reaction time of 2.5 seconds. Thus As shown, if x = 60 mph, then Note that this is the distance traveled before the brake actually begins stopping the vehicle. 8 10/11/2013 Composite Functions Composite Functions 10/11/2013

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**Composition of Functions**

Funcitions of Functions 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 Stopping a Vehicle We note that for a vehicle traveling x miles per hour (mi/hr or mph), stopping distance s(x) is composed of two components: reaction distance r(x), the distance traveled while the driver reacts to a need to stop and hits the brake, and the braking distance b(x), the distance traveled while the brakes are actually applied. Test trials indicate the average human reaction time is about 2.5 seconds. As part of the overall stopping distance calculation, we want to compute the reaction distance, expressed as a function of the speed of the vehicle given in miles per hour. Since we are dealing with reaction times in seconds and want to know how many feet the vehicle travels (not how many miles), we must apply conversion factors to convert miles to feet and hours to seconds. The conversion factors simplify to the constant 11/3 which converts velocity x in mph to feet traveled in the reaction time of 2.5 seconds. Thus As shown, if x = 60 mph, then Note that this is the distance traveled before the brake actually begins stopping the vehicle. This is Example 1 in Section 5.1 of College Algebra by Gary Rockswold. 9 10/11/2013 Composite Functions Composite Functions 10/11/2013

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**Composition of Functions**

Funcitions of Functions 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 Stopping a Vehicle We note that for a vehicle traveling x miles per hour (mi/hr or mph), stopping distance s(x) is composed of two components: reaction distance r(x), the distance traveled while the driver reacts to a need to stop and hits the brake, and the braking distance b(x), the distance traveled while the brakes are actually applied. Test trials indicate the average human reaction time is about 2.5 seconds. As part of the overall stopping distance calculation, we want to compute the reaction distance, expressed as a function of the speed of the vehicle given in miles per hour. Since we are dealing with reaction times in seconds and want to know how many feet the vehicle travels (not how many miles), we must apply conversion factors to convert miles to feet and hours to seconds. The conversion factors simplify to the constant 11/3 which converts velocity x in mph to feet traveled in the reaction time of 2.5 seconds. Thus As shown, if x = 60 mph, then Note that this is the distance traveled before the brake actually begins stopping the vehicle. This is Example 1 in Section 5.1 of College Algebra by Gary Rockswold. 10 10/11/2013 Composite Functions Composite Functions 10/11/2013

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**Composition of Functions**

Funcitions of Functions 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 h g f Stopping a Vehicle We note that for a vehicle traveling x miles per hour (mi/hr or mph), stopping distance s(x) is composed of two components: reaction distance r(x), the distance traveled while the driver reacts to a need to stop and hits the brake, and the braking distance b(x), the distance traveled while the brakes are actually applied. Test trials indicate the average human reaction time is about 2.5 seconds. As part of the overall stopping distance calculation, we want to compute the reaction distance, expressed as a function of the speed of the vehicle given in miles per hour. Since we are dealing with reaction times in seconds and want to know how many feet the vehicle travels (not how many miles), we must apply conversion factors to convert miles to feet and hours to seconds. The conversion factors simplify to the constant 11/3 which converts velocity x in mph to feet traveled in the reaction time of 2.5 seconds. Thus As shown, if x = 60 mph, then Note that this is the distance traveled before the brake actually begins stopping the vehicle. This is Example 1 in Section 5.1 of College Algebra by Gary Rockswold. ● ● ● ● 60 316800 88 220 r(x) x mi/hr y ft/hr z ft/sec feet Composite Functions 11 10/11/2013 Composite Functions 10/11/2013

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**Composition of Functions**

Funcitions of Functions Composite Diagram: (version 1) g f Dom f Range f Range g a f(a) f g ● ● ● x f(x) g(f(x)) g(b) Dom g Composition of Functions The example shows a function f(x) defined on its domain (dark blue set) with range (dark red set) intersecting the domain of function g (yellow set). So function g is defined on a domain overlapping the range of f. Clearly Range f and Dom g are not equal, since there could be an element b in Dom g (the yellow set) which is not in Range f (the red set); that is, there is no x in Dom f (dark blue set) that maps to b in Dom g (yellow set). In this case Dom g is not “contained” in the range of f, that is, Dom g Range f. On the other hand, there might be an element a in Dom f (dark blue set) that does not map into Dom g (yellow set) that is, f(a) Dom g, or in terms of the sets, Range f Dom g. The range of g is independent of that of the other domains and ranges. It may contain images originating in Dom f which were “forwarded” to Range g by the function g, and it may also contain images mapped by g from Dom g, but which are not in Range f. We distinguish these two types of images under g by showing the range of the composite function as the light green set contained in Range g (dark green set). We can now easily see that the domain of the composite function is the light blue subset of Dom f (dark blue set) that maps into Dom g. Remember: the Dom is a subset of the domain of f the Range is a subset of the range of g b g f Dom g f Range g f Question: Where are Dom g f and Range g f ? a Dom f but f(a) Dom g Is Dom g = Range f ? f(x) ≠ b for any x but g(b) Range g So ( )(a) = g(f(a)) does not exist g f … not necessarily ! b Dom g but b Range f Remember: Dom is a subset of Dom f g f Range is a subset of Range g g f Composite Functions 12 10/11/2013 Composite Functions 10/11/2013

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**Composition of Functions**

Funcitions of Functions Composite Diagram (version 2) g f Domain f Range f Range g a f(a) f g ● ● ● x f(x) g(f(x)) b g(b) Domain g Composition of Functions The example shows a function f(x) defined with its range overlapping the domain of function g. The domain of g, while it overlaps the range of f, is not necessarily “contained” in the range of f, nor does contain the range of f. The range of g is independent of the other domains and ranges. The domain of the composite function , that is, Dom , is thus the subset of the Dom f that maps into the domain of g, Dom g. As shown, there may be one or more elements a in the domain of f that do not map into the domain of g. That is, f(a) is not in the domain of g. Is it possible that the domain of g contains an element b not produced by f? The diagram shows such an element b in the domain of g but outside the range of f. In this case the range of f is not a subset of Dom g. The point to remember here is that the domain of is a subset of the domain of f the range of is a subset of the range of g Note to remember: The reason for talking about domains and ranges being subsets is that two sets are equal if and only if each is a subset of the other. Thus if Dom g is to be equal to Range f, then each must be a subset of the other, and any failure in this regard is a demonstration that Range f ≠ Dom g. Dom g f g f Range g f Dom is a subset of Dom f g f Range is a subset of Range g g f 10/11/2013 Composite Functions 13 Composite Functions 10/11/2013

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**Composition of Functions**

Funcitions of Functions Example: f(x) = x + 2 g(x) = 5x + x Dom f = R Dom g = { x x ≥ 0 } Range f = R Range g = { x x ≥ 0 } (g f)(x) = g(f(x)) = g(x + 2) Composition of Functions The example shows a function f(x) apparently defined on the set of all real numbers whose range is also the set of all real numbers. The function g is defined over the set of non-negative real numbers, so the domain of g is a subset of the range of f. The range of g is also the set of non-negative real numbers, since the principal root of x is positive and no negative values of x are allowed. The domain of the composite is thus the subset of the domain of f that maps into the domain of g. Is it possible that the domain of g contains an element y not produced by f? If we now reveal that the expression for the f(x) shown is just a simplification of then we can easily see that x = 2 is not in the domain of f. Hence, 4 is not in the range of f, therefore not available for g to act upon. Clearly 4 is in the domain of g, since g(4) = 5(4) + 41/2 = 22. But, since f cannot produce 4 for g to act on, then 2 is not in the domain of and 22 is not in its range. We also note that is defined as shown and that is defined in similar fashion. Note that, in general, = 5(x + 2) + x + 2 Composite Functions 14 10/11/2013 Composite Functions 10/11/2013

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**Composition of Functions**

Funcitions of Functions Example: f(x) = x + 2 g(x) = 5x + x = 5(x + 2) + x + 2 (g f)(x) Dom g f = { x x ≥ –2 } = [ –2 , ) Range g f = { x x ≥ 0} = [ 0 , ) Composition of Functions The example shows a function f(x) apparently defined on the set of all real numbers whose range is also the set of all real numbers. The function g is defined over the set of non-negative real numbers, so the domain of g is a subset of the range of f. The range of g is also the set of non-negative real numbers, since the principal root of x is positive and no negative values of x are allowed. The domain of the composite is thus the subset of the domain of f that maps into the domain of g. Is it possible that the domain of g contains an element y not produced by f? If we now reveal that the expression for the f(x) shown is just a simplification of then we can easily see that x = 2 is not in the domain of f. Hence, 4 is not in the range of f, therefore not available for g to act upon. Clearly 4 is in the domain of g, since g(4) = 5(4) + 41/2 = 22. But, since f cannot produce 4 for g to act on, then 2 is not in the domain of and 22 is not in its range. We also note that is defined as shown and that is defined in similar fashion. Note that, in general, 10/11/2013 Composite Functions 15 Composite Functions 10/11/2013

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**Composition of Functions**

Funcitions of Functions Example: f(x) = x + 2 g(x) = 5x + x (g f)(x) = 5(x + 2) + x + 2 Question: How do g f and differ ? f g (f g)(x) = f(g(x)) Composition of Functions The example shows a function f(x) apparently defined on the set of all real numbers whose range is also the set of all real numbers. The function g is defined over the set of non-negative real numbers, so the domain of g is a subset of the range of f. The range of g is also the set of non-negative real numbers, since the principal root of x is positive and no negative values of x are allowed. The domain of the composite is thus the subset of the domain of f that maps into the domain of g. Is it possible that the domain of g contains an element y not produced by f? If we now reveal that the expression for the f(x) shown is just a simplification of then we can easily see that x = 2 is not in the domain of f. Hence, 4 is not in the range of f, therefore not available for g to act upon. Clearly 4 is in the domain of g, since g(4) = 5(4) + 41/2 = 22. But, since f cannot produce 4 for g to act on, then 2 is not in the domain of and 22 is not in its range. We also note that is defined as shown and that is defined in similar fashion. Note that, in general, = f ( 5x + x ) = 5x + x + 2 Composite Functions 16 10/11/2013 Composite Functions 10/11/2013

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**Composition of Functions**

Funcitions of Functions Example: f(x) = x + 2 g(x) = 5x + x (g f)(x) = 5(x + 2) + x + 2 = 5x + x + 2 (f g)(x) = [ –2 , ) Dom g f = [ 0 , ) Range g f Dom f g = [ 0 , ) = [ 2 , ) Range f g Composition of Functions The example shows a function f(x) apparently defined on the set of all real numbers whose range is also the set of all real numbers. The function g is defined over the set of non-negative real numbers, so the domain of g is a subset of the range of f. The range of g is also the set of non-negative real numbers, since the principal root of x is positive and no negative values of x are allowed. The domain of the composite is thus the subset of the domain of f that maps into the domain of g. Is it possible that the domain of g contains an element y not produced by f? If we now reveal that the expression for the f(x) shown is just a simplification of then we can easily see that x = 2 is not in the domain of f. Hence, 4 is not in the range of f, therefore not available for g to act upon. Clearly 4 is in the domain of g, since g(4) = 5(4) + 41/2 = 22. But, since f cannot produce 4 for g to act on, then 2 is not in the domain of and 22 is not in its range. We also note that is defined as shown and that is defined in similar fashion. Note that, in general, What if f(x) = x2 – 4 x – 2 = x + 2 , for x ≠ 2 ? 10/11/2013 Composite Functions 17 Composite Functions 10/11/2013

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**Composition of Functions**

Funcitions of Functions Example: f(x) = x + 2 g(x) = 5x + x f(x) = x2 – 4 x – 2 = x + 2 , for x ≠ 2 ? = { x | x ≥ –2 , x ≠ 2 } Dom g f = { x | x ≥ 0 , x ≠ 22 } Range g f Composition of Functions The example shows a function f(x) apparently defined on the set of all real numbers whose range is also the set of all real numbers. The function g is defined over the set of non-negative real numbers, so the domain of g is a subset of the range of f. The range of g is also the set of non-negative real numbers, since the principal root of x is positive and no negative values of x are allowed. The domain of the composite is thus the subset of the domain of f that maps into the domain of g. Is it possible that the domain of g contains an element y not produced by f? If we now reveal that the expression for the f(x) shown is just a simplification of then we can easily see that x = 2 is not in the domain of f. Hence, 4 is not in the range of f, therefore not available for g to act upon. Clearly 4 is in the domain of g, since g(4) = 5(4) + 41/2 = 22. But, since f cannot produce 4 for g to act on, then 2 is not in the domain of and 22 is not in its range. We also note that is defined in similar fashion. Since g(x) is defined for all positive x, ( )(x) is defined for all x for which g(x) ≠ 2, that is for 5x + x1/2 ≠ 2, or approximately, x ≠ With g(x) ≠ 2 then ( )(x) ≠ /2 ≈ . Dom f g = { x | x ≥ 0 , x ≠ } Range f g = { x | x ≥ 2, x ≠ } 10/11/2013 Composite Functions 18 Composite Functions 10/11/2013

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**Composition of Functions**

Funcitions of Functions f(x) = x2 – 4 x – 2 Example: Composite Diagram = x + 2 , for x ≠ 2, f(x) ≠ 4 g(x) = 5x + x (g f)(x) = g(f(x)) = g(x + 2) = 5(x + 2) + x + 2 f Domain f Range f Range g –5 –3 g f Composition of Functions The example shows a function f(x) apparently defined on the set of all real numbers whose range is also the set of all real numbers. The function g is defined over the set of non-negative real numbers, so the domain of g is a subset of the range of f. The range of g is also the set of non-negative real numbers, since the principal root of x is positive and no negative values of x are allowed. The domain of the composite is thus the subset of the domain of f that maps into the domain of g. Is it possible that the domain of g contains an element y not produced by f? If we now reveal that the expression for f(x) shown is actually a simplification of then we can easily see that x = 2 is not in the domain of f. Hence, 4 is not in the range of f, therefore not available for g to act upon. Clearly 4 is in the domain of g, since g(4) = 5(4) + 41/2 = 22. We also note that is defined as shown and that is defined in similar fashion. Note that, in general, ● ● ● x f(x) g(f(x)) Range g f = { x x ≥ 0 , x ≠ 22 } Dom g f = { x x ≥ –2 , x ≠ 2 } 4 22 Domain g g f Composite Functions 19 10/11/2013 Composite Functions 10/11/2013

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**Composition of Functions**

Funcitions of Functions Composition Versus Multiplication NOTE: Example : Let f(x) = x2 and g(x) = x1/2 (g f)(x) ≠ (gf)(x) ... then (g f)(x) = g(f(x)) = g(x2) = (x2)1/2 = |x| , for all x Composition Versus Multiplication The example shows that composition and multiplication are NOT equivalent. That is, composition is NOT multiplication !! One way to remember this is to observe that for multiplication f(x) and g(x) must be evaluated independently and simultaneously prior to multiplication. For composition of g with f, or f with g, we evaluate one of the functions first (the so-called inner function) and use its output as input to the second (or outer) function. Thus evaluation is done consecutively not simultaneously. Can = ? For certain values of x (namely non-negative real numbers) the functional values are the same. Question is, are the two functions really the same function? The answer depends on their domains: they are the same function if and only if their domains are equal ! Are they equal? Note that Dom f = R and Dom g = { x x ≥ 0 }. Dom = R , but Dom = { x x ≥ 0 }. Thus, can operate on all real numbers, but can operate only on non-negative real numbers. Thus and are not the same function ! But … (gf)(x) = g(x) • f(x) = x1/2 • x2 = x5/2 ... which is NOT |x| !! But … Composite Functions 20 10/11/2013 Composite Functions 10/11/2013

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**Composition of Functions**

Funcitions of Functions Composition Versus Multiplication Is it possible that ? (g f)(x) = (f g)(x) ... well ... (f g)(x) = f(g(x)) = f(x1/2) = (x1/2)2 = x Composition Versus Multiplication The example shows that composition and multiplication are NOT equivalent. That is, composition is NOT multiplication !! One way to remember this is to observe that for multiplication f(x) and g(x) must be evaluated independently and simultaneously prior to multiplication. For composition of g with f, or f with g, we evaluate one of the functions first (the so-called inner function) and use its output as input to the second (or outer) function. Thus evaluation is done consecutively not simultaneously. Can = ? For certain values of x (namely non-negative real numbers) the functional values are the same. Question is, are the two functions really the same function? The answer depends on their domains: they are the same function if and only if their domains are equal ! Are they equal? Note that Dom f = R and Dom g = { x x ≥ 0 }. Dom = R , but Dom = { x x ≥ 0 }. Thus, can operate on all real numbers, but can operate only on non-negative real numbers. Thus and are not the same function ! = (g f)(x) , for x ≥ 0 Question: Are and the same function ? g f f g Only if their domains are equal ! Are they ? Composite Functions 21 10/11/2013 Composite Functions 10/11/2013

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**Funcitions of Functions**

Combining Functions Funcitions of Functions Find and graphically (g f)(–3) (f g)(2) -3 -1 -2 3 1 2 x y -3 -1 -2 3 1 2 x y f(1) = 3 ● ● g(2) = 1 ● ● ● ● f(-3) = -1 ● ● y = f(x) y =g(x) y = f(x) y =g(x) Composition Examples: Graphically We can use the graph of the inner (first) function to determine the input to the outer (second) function. The output of the first function appears on the y-axis, so it has to be transferred (the black arrows) to the x-axis for input to the second function. Notice that the two composite functions are not the same function. = f(g(2)) (f g)(2) = g(f(-3)) (g f)(-3) = f(1) = 3 = -2 = g(-1) Note: Output of the first function is input to the second Composite Functions 22 10/11/2013 Composite Functions 10/11/2013

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**Funcitions of Functions**

Combining Functions Funcitions of Functions Find and in tabular form (g f)(–3) (f g)(2) –3 –1 x –3 –2 – 1 2 –1 3 f(x) – –2 1 g(x) –2 –3 – (f g)(2) = f(g(2)) = f(1) = 3 Composition Examples: Tabular Form Using a table of selected values (since we cannot list all the ordered pairs for the functions) we can find values of the composite functions from the previous slide. As before, the thing to remember is that the output from the inner (first) function is input to the outer (second) function. (g f)(-3) = g(f(-3)) = g(-1) = -2 Note: Output of the first function is input to the second 10/11/2013 Composite Functions 23 Composite Functions 10/11/2013

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**Funcitions of Functions**

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

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