Stitz-Zeager_College_Algebra_e-book

Stitz-Zeager_College_Algebra_e-book

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Unformatted text preview: = g (f (x)) = g x2 − 4x = 2 − Hence, (g ◦ f )(x) = 2 − √ x2 − 4x + 3. (x2 − 4x) + 3 = 2 − x2 − 4x + 3 5.1 Function Composition 281 • outside in : We use the formula for g first to get (x2 − 4x) + 3 = 2 − √ We get the same answer as before, (g ◦ f )(x) = 2 − x2 − 4x + 3. (g ◦ f )(x) = g (f (x)) = 2 − f (x) + 3 = 2 − x2 − 4x + 3 To find the domain of g ◦ f , we need to find the elements in the domain of f whose outputs f (x) are in the domain of g . We accomplish this by following the rule set forth in Section 1.5, that is, we find the domain before we simplify. To that end, we examine (g ◦ f )(x) = 2 − (x2 − 4x) + 3. To keep the square root happy, we solve the inequality x2 − 4x + 3 ≥ 0 by creating a sign diagram. If we let r(x) = x2 − 4x + 3, we find the zeros of r to be x = 1 and x = 3. We obtain (+) 0 (−) 0 (+) 1 3 Our solution to x2 − 4x + 3 ≥ 0, and hence the domain of g ◦ f , is (−∞, 1] ∪ [3, ∞). 2. To find (f ◦ g )(x), we find f (g (x)). • in...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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