24_Cal_Solution of Calculus_6e

24_Cal_Solution of Calculus_6e - x √ x − 2 = p x 2 −...

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Section 1.3 C01S03.001: The domain of f is R , the set of all real numbers; so is the domain of g , but g ( x ) = 0 when x = 1 and when x = 3. So the domain of f + g and f · g is the set R and the domain of f/g is the set of all real numbers other than 1 and 3. Their formulas are ( f + g )( x )= x 2 +3 x 2 , ( f · g )( x )=( x + 1)( x 2 +2 x 3) = x 3 +3 x 2 x 3 , and µ f g ( x )= x +1 x 2 +2 x 3 . C01S03.002: The domain of f consists of all real numbers other than 1 and the domain of g consists of all real numbers other than 1 2 . Hence the domain of f + g , f · g , and f/g consists of all real numbers other than 1 2 and 1. For such x , ( f + g )( x )= 1 x 1 + 1 2 x +1 = 3 x ( x 1)(2 x +1) , ( f · g )( x )= 1 ( x 1)(2 x +1) , and µ f g ( x )= 2 x +1 x 1 . Note that, in spite of the last equation, the domain of f/g does not include the number 1 2 . C01S03.003: The domain of f is the interval [0 , + ) and the domain of g is the interval [2 , + ). Hence the domain of f + g and f · g is the interval [2 , + ), but because g (2) = 0, the domain of f/g is the open interval (2 , + ). The formulas for these combinations are ( f + g )( x )= x + x 2 , ( f · g )( x )=
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Unformatted text preview: x √ x − 2 = p x 2 − 2 x, and µ f g ¶ ( x ) = √ x √ x − 2 = r x x − 2 . C01S03.004: The domain of f is the interval [ − 1 , + ∞ ) and the domain of g is the interval ( −∞ , 5]. Hence the domain of f + g and f · g is the closed interval [ − 1 , 5], but because g (5) = 0, the domain of f/g is the half-open interval [ − 1 , 5). Their formulas are ( f + g )( x ) = √ x + 1 + √ 5 − x, ( f · g )( x ) = √ x + 1 √ 5 − x = p 5 + 4 x − x 2 , and µ f g ¶ ( x ) = √ x + 1 √ 5 − x = r x + 1 5 − x . C01S03.005: The domain of f is the set R of all real numbers; the domain of g is the open interval ( − 2 , 2). Hence the domain of f + g and f · g is the open interval ( − 2 , 2); because g ( x ) is never zero, the domain of f/g is the same. Their formulas are 1...
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