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Chapter 2: More on Functions b) (f + g)(x) = f (x) + g(x) = (2x − 1) + (−2x2 ) = −2x2 + 2x − 1 (f − g)(x) = f (x) − g(x) = (2x − 1) − (−2x2 ) = 2x2 + 2x − 1 (f g)(x) = f (x) · g(x) = (2x − 1)(−2x2 ) = −4x3 + 2x2 (f f )(x) = f (x) · f (x) = (2x − 1)(2x − 1) = 4x2 − 4x + 1 (f /g)(x) =
2x − 1 f (x) = g(x) −2x2
(g/f )(x) =
−2x2 g(x) = f (x) 2x − 1
22. f (x) = x2 − 1, g(x) = 2x + 5 a) The domain of f and of g is the set of all real numbers, or (−∞, ∞). Then the domain of f + g, f − g, 5 = 0, the f g and f f is (−∞, ∞). Since g − 2
5 5 ∪ − , ∞ . Since domain of f /g is − ∞, − 2 2 f (1) = 0 and f (−1) = 0, the domain of g/f is (−∞, −1) ∪ (−1, 1) ∪ (1, ∞).
24. f (x) =
√
x, g(x) =
√
2−x
a) The domain of f is [0, ∞). The domain of g is (−∞, 2]. Then the domain of f + g, f − g, and f g is [0, 2]. The domain of f f is the same as the domain of f , [0, ∞). Since g(2) = 0, the domain of f /g is [0, 2). Since f (0) = 0, the domain of g/f is (0, 2]. √ √ b) (f + g)(x) = x + 2 − x √ √ (f − g)(x) = x − 2 − x √ √ √ (f g)(x) = x · 2 − x = 2x − x2 √ √ √ (f f )(x) = x · x = x2 = |x| √ x (f /g)(x) = √ 2−x √ 2−x (g/f )(x) = √ x 25. f (x) = x + 1, g(x) = |x|
(f − g)(x) = x2 − 1 − (2x + 5) = x2 − 2x − 6
a) The domain of f and of g is (−∞, ∞). Then the domain of f + g, f − g, f g, and f f is (−∞, ∞). For f /g, we must exclude 0 since g(0) = 0. The domain of f /g is (−∞, 0) ∪ (0, ∞). For g/f , we must exclude −1 since f (−1) = 0. The domain of g/f is (−∞, −1) ∪ (−1, ∞).
(f g)(x) = (x2 −1)(2x+5) = 2x3 +5x2 −2x−5
b) (f + g)(x) = f (x) + g(x) = x + 1 + |x|
b) (f + g)(x) = x2 − 1 + 2x + 5 = x2 + 2x + 4
(f − g)(x) = f (x) − g(x) = x + 1 − |x|
(f f )(x) = (x − 1) = x − 2x + 1 2
2
4
2
(f g)(x) = f (x) · g(x) = (x + 1)|x|
x2 − 1 2x + 5 2x + 5 (g/f )(x) = 2 x −1 √ √ 23. f (x) = x − 3, g(x) = x + 3 (f /g)(x) =
(f f )(x) = f (x)·f (x) = (x+1)(x+1) = x2 + 2x + 1 x+1 (f /g)(x) = |x| (g/f )(x) =
a) Since f (x) is nonnegative for values of x in [3, ∞), this is the domain of f . Since g(x) is nonnegative for values of x in [−3, ∞), this is the domain of g. The domain of f +g, f −g, and f g is the intersection of the domains of f and g, or [3, ∞). The domain of f f is the same as the domain of f , or [3, ∞). For f /g, we must exclude −3 since g(−3) = 0. This is not in [3, ∞), so the domain of f /g is [3, ∞). For g/f , we must exclude 3 since f (3) = 0. The domain of g/f is (3, ∞). √ √ b) (f + g)(x) = f (x) + g(x) = x − 3 + x + 3 √ √ (f − g)(x) = f (x) − g(x) = x − 3 − x + 3 √ √ √ (f g)(x) = f (x) · g(x) = x−3 · x + 3 = x2 −9 √ √ (f f )(x) = f (x) · f (x) = x − 3 · x − 3 = |x − 3| √ x−3 (f /g)(x) = √ x+3 √ x+3 (g/f )(x) = √ x−3
Copyright
|x| x+1
26. f (x) = 4|x|, g(x) = 1 − x a) The domain of f and of g is (−∞, ∞). Then the domain of f +g, f −g, f g, and f f is (−∞, ∞). Since g(1) = 0, the domain of f /g is (−∞, 1) ∪ (1, ∞). Since f (0) = 0, the domain of g/f is (−∞, 0) ∪ (0, ∞). b) (f + g)(x) = 4|x| + 1 − x (f − g)(x) = 4|x| − (1 − x) = 4|x| − 1 + x (f g)(x) = 4|x|(1 − x) = 4|x| − 4x|x| (f f )(x) = 4|x| · 4|x| = 16x2 4|x| 1−x 1−x (g/f )(x) = 4|x| (f /g)(x) =
27. f (x) = x3 , g(x) = 2x2 + 5x − 3 a) Since any number can be an input for either f or g, the domain of f , g, f + g, f − g, f g, and f f is the set of all real numbers, or (−∞, ∞). 1 = 0, the domain of f /g Since g(−3) = 0 and g 2
1 1 ∪ ,∞ . is (−∞, −3) ∪ − 3, 2 2
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