25_Cal_Solution of Calculus_6e

25_Cal_Solution of Calculus_6e - ( f + g )( x ) = p x 2 + 1...

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Unformatted text preview: ( f + g )( x ) = p x 2 + 1 + 1 √ 4 − x 2 , ( f · g )( x ) = √ x 2 + 1 √ 4 − x 2 , and µ f g ¶ ( x ) = p x 2 + 1 p 4 − x 2 = p 4 + 3 x 2 − x 4 . C01S03.006: The domain of f is the set of all real numbers other than 2 and the domain of g is the set of all real numbers other than − 2. Hence the domain of f + g and f · g is the set of all real numbers other than ± 2. But because g ( − 1) = 0, − 1 does not belong to the domain of f/g , which therefore consists of all real numbers other than − 2, − 1, and 2. The formulas of these combinations are ( f + g )( x ) = x − 1 x − 2 + x + 1 x + 2 = 2 x 2 − 4 x 2 − 4 , ( f · g )( x ) = x − 1 x − 2 · x + 1 x + 2 = x 2 − 1 x 2 − 4 , and µ f g ¶ ( x ) = x − 1 x − 2 · x + 2 x + 1 = x 2 + x − 2 x 2 − x − 2 . C01S03.007: f ( x ) = x 3 − 3 x + 1 has 1, 2, or 3 zeros, approaches + ∞ as x does, and approaches −∞ as x does. Because f (0) 6 = 0, the graph does not match Fig. 1.3.26, so it must match Fig. 1.3.30.= 0, the graph does not match Fig....
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This note was uploaded on 09/14/2011 for the course MATH 101 taught by Professor Cheng during the Spring '11 term at Ecole Hôtelière de Lausanne.

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