Stitz-Zeager_College_Algebra_e-book

# We move the non zero terms to the left leave a 0 on

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Unformatted text preview: 3 x+1 5x+3 x+1 . Hence, (g ◦ h)(x) = 2 − simplify: (g ◦ h)(x) = 2 − +3 get common denominators as before To ﬁnd the domain, we look to the step before we began to 2x x+1 + 3. To avoid division by zero, we need x = −1. To keep the radical happy, we need to solve x2x + 3 ≥ 0. Getting common denominators as before, +1 x+3 x+3 this reduces to 5x+1 ≥ 0. Deﬁning r(x) = 5x+1 , we have that r is undeﬁned at x = −1 and 3 r(x) = 0 at x = − 5 . We get (+) (−) 0 (+) −1 −3 5 3 Our domain is (−∞, −1) ∪ − 5 , ∞ . 4. We ﬁnd (h ◦ g )(x) by ﬁnding h(g (x)). • inside out : We insert the expression g (x) into h ﬁrst to get (h ◦ g )(x) = h(g (x)) 5.1 Function Composition 283 = = = √ h 2− x+3 √ 2 2− x+3 √ 2− x+3 +1 √ 4−2 x+3 √ 3− x+3 • outside in : We use the formula for h(x) ﬁrst to get (h ◦ g )(x) = = = = h(g (x)) 2 (g (x)) (g (x)) + 1 √ 2 2− x+3 √ 2− x+3 +1 √ 4−2 x+3 √ 3− x+3 √ 4−2 x+3 √ . To ﬁnd the 3− x+3 √ 2(2− x+3)...
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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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