Stitz-Zeager_College_Algebra_e-book

F c 1 2c 5 1 2c 2c c f d 1 2d 5 1 2d 2d d hence

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Unformatted text preview: side out : We insert the expression g (x) into f first to get (f ◦ g )(x) = = = = = = f (g (x)) √ f 2− x+3 √ √ 2 2− x+3 −4 2− x+3 √ √ √ 2 4−4 x+3+ x+3 −8+4 x+3 4+x+3−8 x−1 • outside in : We use the formula for f (x) first to get (f ◦ g )(x) = = = = f (g (x)) (g (x))2 − 4 (g (x)) √ √ 2 2− x+3 −4 2− x+3 x−1 same algebra as before Thus we get (f ◦ g )(x) = x − 1. To find the domain of (f ◦ g ), we look to the step before √ √ 2 we did any simplification and find (f ◦ g )(x) = 2 − x + 3 − 4 2 − x + 3 . To keep the square root happy, we set x + 3 ≥ 0 and find our domain to be [−3, ∞). 3. To find (g ◦ h)(x), we compute g (h(x)). 282 Further Topics in Functions • inside out : We insert the expression h(x) into g first to get (g ◦ h)(x) = g (h(x)) = g = 2− = 2− 2x 3(x + 1) + x+1 x+1 = 2− 5x + 3 x+1 2x x+1 2x x+1 +3 get common denominators • outside in : We use the formula for g (x) first to get (g ◦ h)(x) = = g (h(x)) 2 − h(x) + 3 = 2− 2x x+1 = 2− 5x +...
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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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