Stitz-Zeager_College_Algebra_e-book

1 although we discussed imaginary numbers in section

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Unformatted text preview: (−∞, ∞) (c) (g ◦ f )(x) = −x2 Domain: [−3, 3] √ (d) (f ◦ g )(x) =√ −x4 √ 18x2√ 72 + −√ Domain: [− 12, − 6] ∪ [ 6, 12]6 5. (a) (f ◦ g )(3) = f (g (3)) = f (2) = 4 (h) (g ◦ f )(−2) = g (f (−2)) = g (2) = 1 4. (a) (f ◦ f )(x) = |x| Domain: [−3, 3] (b) f (g (−1)) = f (−4) which is undefined (i) g (f (g (0))) = g (f (0)) = g (1) = −3 (c) (f ◦ f )(0) = f (f (0)) = f (1) = 3 (j) f (f (f (−1))) = f (f (0)) = f (1) = 3 (d) (f ◦ g )(−3) = f (g (−3)) = f (−2) = 2 (f) g (f (−3)) = g (4) which is undefined (k) f (f (f (f (f (1))))) = f (f (f (f (3)))) = f (f (f (−1))) = f (f (0)) = f (1) = 3 n times (g) (g ◦ g )(−2) = g (g (−2)) = g (0) = 0 (l) (g ◦ g ◦ · · · ◦ g )(0) = 0 (e) (g ◦ f )(3) = g (f (3)) = g (−1) = −4 √ 6. F (x) = 3 −x + 2 − 4 = k (j (f (h(g (x))))) 1 7. One possible solution is F (x) = − 2 (2x − 7)3 + 1 = k (j (f (h(g (x))))) where g (x) = 2x, h(x) = 1 x − 7, j (x) = − 2 x and k (x) = x + 1. You could also have F...
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