Quiz4sol - P ( x ): P (1) = (1) 3-2(1) 2-5(1) + 6 = 0....

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MATH 1450-01: Algebra and Trigonometry, Fall 2010 Quiz 4 September 27, 2010 Answer Key 1. (a) ( f + g )(1) = f (1) + g (1) = - 1 + 1 = 0 (b) f g ! (1) = f (1) g (1) = - 1 1 = - 1 (c) ( f og )(1) = f ( g (1)) = f (1) = - 1 (d) ( f g )(1) = f (1) × g (1) = 1 × - 1 = - 1 2. (a) f ( x ) = x 2 - 1. The graph of f ( x ) is a translation of that of x 2 and we know the graph of x 2 fails the horizontal line test. Therefore f ( x ) is not one-to-one. Therefore it has no inverse function. (b) g ( x ) = x - 1. The graph of g ( x ) is a translation of that of x and we know the graph of x passes the horizontal line test. Therefore g ( x ) is one-to-one and so has an inverse function. Let y = x - 1. Then y 2 = x - 1 = x = y 2 + 1 = g - 1 ( y ) = y 2 + 1 = g - 1 ( x ) = x 2 + 1. 3. (a) The factors of a 0 = 6 are p = ± 1 , ± 2 , ± 3 , ± 6, and the factors of a n = 1 are q = ± 1. Then by the Rational zero theorem, the possible zeros are p q = ± 1 , ± 2 , ± 3 , ± 6 . (b) We test all the possible zeros of
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Unformatted text preview: P ( x ): P (1) = (1) 3-2(1) 2-5(1) + 6 = 0. Therefore 1 is a zero of the polynomial. P (-1) = (-1) 3-2(-1) 2-5(-1) + 6 , 0. P (2) = (2) 3-2(2) 2-5(2) + 6 , 0. P (-2) = (-2) 3-2(-2) 2-5(-2) + 6 = 0. Therefore -2 is a zero of the polynomial. P (3) = (3) 3-2(3) 2-5(3) + 6 = 0. Therefore 3 is a zero of the polynomial. Since p ( x ) is a polynomial of degree it has at most three zeros. So the zeros are 1 ,-2 ,-3. (c) As x → + ∞ , x 3 → + ∞ = ⇒ P ( x ) → + ∞ . As x → -∞ , x 3 → -∞ = ⇒ P ( x ) → -∞ . (d) . 1...
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This note was uploaded on 09/30/2010 for the course MATH 1450 taught by Professor A during the Spring '10 term at Zhejiang University.

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