Stitz-Zeager_College_Algebra_e-book

Theorem 52 tells us equation y f 1 x is equivalent to

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Unformatted text preview: before 2 To find the domain of h ◦ h, we analyze (h ◦ h)(x) = 2x x+1 2x x+1 x + 1 happy, we need x = −1. Setting the denominator 1 domain is (−∞, −1) ∪ −1, − 3 ∪ − 1 , ∞ . 3 . To keep the denominator +1 2x x+1 + 1 = 0 gives x = − 1 . Our 3 6. The expression (h ◦ (g ◦ f ))(x) indicates that we first find the composite, g ◦ f√ and compose the function h with the result. We know from number 1 that (g ◦ f )(x) = 2 − x2 − 4x + 3. We now proceed as usual. • inside out : We insert the expression (g ◦ f )(x) into h first to get (h ◦ (g ◦ f ))(x) = = = h((g ◦ f )(x)) √ h 2 − x2 − 4x + 3 √ 2 2 − x2 − 4x + 3 √ 2 − x2 − 4x + 3 + 1 5.1 Function Composition 285 = √ 4 − 2 x2 − 4x + 3 √ 3 − x2 − 4x + 3 • outside in : We use the formula for h(x) first to get (h ◦ (g ◦ f ))(x) = = = = h((g ◦ f )(x)) 2 ((g ◦ f )(x)) ((g ◦ f )(x)) + 1 √ 2 2 − x2 − 4x + 3 √ 2 − x2 − 4x + 3 + 1 √ 4 − 2 x2 − 4x + 3 √ 3 − x2 − 4x + 3 √ 4−2 x2 −4x+3 √ . 3− x2 −4x+3 To find the domain, we look at the step before √ 2(2− x2 −4x+3) we began to simplify, (h ◦ (g ◦ f...
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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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