276 CHAPTER 5 Exponential and Logarithmic Functions Seeing the Concept Using a graphing calculator, let Y f x x Y g x x Y f g Y g f 3 4 4 3 , 1 2 3 4 ( ) ( ) = = − = = + = = Using the viewing window − ≤ ≤ x 3 3, − ≤ ≤ y 2 2, graph only Y3 and Y .4 What do you see? TRACE to verify that Y Y . 3 4 = In Example 3, the domain of f g was found to be x x x | 1, 1 . { } ≠ − ≠ The domain of f g also can be found by first looking at the domain of g : x x| 1 . { } ≠ Exclude 1 from the domain of f g as a result. Then look at f g and note that x cannot equal 1, − because x 1 = − results in division by 0. So exclude 1− from the domain of f g. Therefore, the domain of f g is x x x | 1, 1 . { } ≠ − ≠ (b) ( ) ( ) ( ) ( ) ( ) ( ) = = + = + + = + + + = + + f f x f f x f x x x x x x 1 2 1 1 2 2 2 1 2 2 2 2 5 ↑ ↑ ↑ ( ) ( ) = + = + + + f x x f x x x x 1 2 1 2 Multiplyby 2 2 . The domain of f f consists of all values of x in the domain of f x x , | 2 , { } ≠ − for which f x x 1 2 2. ( ) = + ≠ − To find other numbers x to exclude, solve the equation ( ) + =− =− + =− − =− =− x x x x x 1 2 2 1 2 2 1 2 4 2 5 5 2 So f x 2 ( ) ≠ − if ≠ − x 5 2 . The domain of f f is { } ≠ − ≠ − x x x 5 2 , 2 . The domain of f f also can be found by noting that 2− is not in the domain of f and so is not in the domain of f f. Then, looking at f f, note that x cannot equal − 5 2 . Do you see why? Therefore, the domain of f f is { } ≠ − ≠ − x x x 5 2 , 2 . EXAMPLE 5 Showing That Two Composite Functions Are Equal If f x x3 4 ( ) = − and g x x 4 3 , ( ) = + show that f g x g f x x ( ) ( ) ( ) ( ) = = for every x in the domain of f g and g f. Solution f g x f g x f x x x x 4 3 3 4 3 4 4 4 ( ) ( ) ( ) ( ) ( ) ( ) = = + = + − = + − = ↑ ↑ ( ) ( ) = + = − g x x f x x 4 3 3 4 g f x g f x g x x x x 3 4 3 4 4 3 3 3 ( ) ( ) ( ) ( ) ( ) ( ) = = − = − + = = ↑ ↑ ( ) ( ) = − = + f x x g x x 3 4 4 3 We conclude that f g x g f x x x , anyrealnumber. ( ) ( ) ( ) ( ) = = Now Work PROBLEMS 27 AND 29
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