Inside out we insert the expression f x into h g

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inside out : We insert the expression f ( x ) into h g first to get (( h g ) f )( x ) = ( h g )( f ( x )) = ( h g ) ( x 2 - 4 x ) = 4 - 2 p ( x 2 - 4 x ) + 3 3 - p ( x 2 - 4 x ) + 3 = 4 - 2 x 2 - 4 x + 3 3 - x 2 - 4 x + 3 outside in : We use the formula for ( h g )( x ) first to get (( h g ) f )( x ) = ( h g )( f ( x )) = 4 - 2 p ( f ( x )) + 3 3 - p f ( x )) + 3 = 4 - 2 p ( x 2 - 4 x ) + 3 3 - p ( x 2 - 4 x ) + 3 = 4 - 2 x 2 - 4 x + 3 3 - x 2 - 4 x + 3 We note that the formula for (( h g ) f )( x ) before simplification is identical to that of ( h ( g f ))( x ) before we simplified it. Hence, the two functions have the same domain, h ( f g ) is ( -∞ , 2 - 10) (2 - 10 , 1] 3 , 2 + 10 ) ( 2 + 10 , ) . It should be clear from Example 5.1.1 that, in general, when you compose two functions, such as f and g above, the order matters. 4 We found that the functions f g and g f were different as were g h and h g . Thinking of functions as processes, this isn’t all that surprising. If we think of one process as putting on our socks, and the other as putting on our shoes, the order in which we do these two tasks does matter. 5 Also note the importance of finding the domain of the composite function before simplifying. For instance, the domain of f g is much different than its simplified formula would indicate. Composing a function with itself, as in the case of finding ( g g )(6) and ( h h )( x ), may seem odd. Looking at this from a procedural perspective, however, this merely indicates performing a task h and then doing it again - like setting the washing machine to do a ‘double rinse’. Composing a function with itself is called ‘iterating’ the function, and we could easily spend an entire course on just that. The last two problems in Example 5.1.1 serve to demonstrate the associative property of functions. That is, when composing three (or more) functions, as long as we keep the order the same, it doesn’t matter which two functions we compose first. This property as well as another important property are listed in the theorem below. 4 This shows us function composition isn’t commutative . An example of an operation we perform on two functions which is commutative is function addition, which we defined in Section 1.5 . In other words, the functions f + g and g + f are always equal. Which of the remaining operations on functions we have discussed are commutative? 5 A more mathematical example in which the order of two processes matters can be found in Section 1.7 . In fact, all of the transformations in that section can be viewed in terms of composing functions with linear functions.
5.1 Function Composition 405 Theorem 5.1. Properties of Function Composition: Suppose f , g , and h are functions. h ( g f ) = ( h g ) f , provided the composite functions are defined. If I is defined as I ( x ) = x for all real numbers x , then I f = f I = f .

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