300 CHAPTER 2 Graphs and Functions SOLUTION (a) From the figure, repeated in the margin, ƒ142 = 9 and g142 = 2. 1ƒ + g2142 = ƒ142 + g142 1ƒ + g21x2 = ƒ1x2 + g1x2 = 9 + 2 Substitute. = 11 Add. For 1ƒ - g21-22, although ƒ1-22 = -3, g1-22 is undefined because -2 is not in the domain of g. Thus 1ƒ - g21-22 is undefined. The domains of ƒ and g both include 1. 1ƒg2112 = ƒ112 # g112 1ƒg21x2 = ƒ1x2 # g1x2 = 3 # 1 Substitute. = 3 Multiply. The graph of g includes the origin, so g102 = 0. Thus A f gB102 is undefined. (b) From the table, repeated in the margin, ƒ142 = 9 and g142 = 2. 1ƒ + g2142 = ƒ142 + g142 1ƒ + g21x2 = ƒ1x2 + g1x2 = 9 + 2 Substitute. = 11 Add. In the table, g1-22 is undefined, and thus 1ƒ - g21-22 is also undefined. 1ƒg2112 = ƒ112 # g112 1fg21x2 = ƒ1x2 # g1x2 = 3 # 1 ƒ112 = 3 and g112 = 1 = 3 Multiply. The quotient function value A ƒ gB102 is undefined because the denominator, g102, equals 0. 1ƒ + g2142 = ƒ142 + g142 = 12 # 4 + 12 + 24 = 9 + 2 = 11 1ƒg2112 = ƒ112 # g112 = 12 # 1 + 12 # 21 = 3112 = 3 A ƒ gB102 is undefined since g102 = 0. S Now Try Exercises 33 and 37. –4 –2 1 2 4 –3 –1 1 5 9 x y y = f(x) y = g(x) x ƒ1x2 g1x2 -2 -3 undefined 0 1 0 1 3 1 4 9 2 (c) Using ƒ1x2 = 2x + 1 and g1x2 = 2x, we can find 1ƒ + g2142 and 1ƒg2112. Because -2 is not in the domain of g, 1ƒ - g21-22 is not defined.
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