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assignment1 solutions - MATH 137 Assignment #1 Solutions...

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Unformatted text preview: MATH 137 Assignment #1 Solutions Fall 2009 1. | 2 x 3- 4 x 2 + 3 x- 1 | ≤ | 2 x 3 | + | - 4 x 2 | + | 3 x | + | - 1 | by the triangle inequality = 2 | x 3 | + 4 | x 2 | + 3 | x | + 1 ≤ 2 | (- 2) 3 | + 4 | (- 2) 2 | + 3 | - 2 | + 1 on the interval- 2 ≤ x ≤ 1 = 16 + 16 + 6 + 1 = 39 2. a) Notice that when x = 0 , 1 , 2 , 3, we have x ( x- 1)( x- 2)( x- 3) = 0 and thus H ( x ( x- 1)( x- 2)( x- 3)) = 1. So let’s consider the following intervals: x < 0 : then x ( x- 1)( x- 2)( x- 3) > 0, so H ( x ( x- 1)( x- 2)( x- 3)) = 1 < x < 1 : then x > , x- 1 < , x- 2 < , x- 3 < 0, and so x ( x- 1)( x- 2)( x- 3) < 0 and H ( x ( x- 1)( x- 2)( x- 3)) = 0 1 < x < 2 : then x > , x- 1 > , x- 2 < , x- 3 < 0, and so x ( x- 1)( x- 2)( x- 3) > 0 and H ( x ( x- 1)( x- 2)( x- 3)) = 1 2 < x < 3 : then x > , x- 1 > , x- 2 > , x- 3 < 0, and so x ( x- 1)( x- 2)( x- 3) < 0 and H ( x ( x- 1)( x- 2)( x- 3)) = 0 x > 3 : then x > , x- 1 > , x- 2 > , x- 3 > 0, and so x ( x- 1)( x- 2)( x- 3) > 0 and H ( x ( x- 1)( x- 2)( x- 3)) = 1 So we get the following sketch: b) Case 1 : x ≥ In this case, | x | = x, H ( x ) = 1 , and H (- x ) = 0. So we have, LHS = | x | = x RHS = xH ( x )- xH (- x ) = x · 1- x · 0 = x = LHS, as required. Case 2 : x < In this case, | x | =- x, H ( x ) = 0 , and H (- x ) = 1. So we have, LHS = | x | =- x RHS = xH ( x )- xH (- x ) = x ·- x · 1 =- x = LHS, as required. c) Case 1 : x <- 1 In this case, H ( x + 1) = 0 and H ( x- 1) = 0 Therefore, y = (1- x 2 )(0- 0) = 0 Case 2 :- 1 ≤ x < 1 In this case, H ( x + 1) = 1 and H ( x- 1) = 0 Therefore, y = (1- x 2 )(1- 0) = 1- x 2 Case 3 : x ≥ 1 In this case, H ( x + 1) = 1 and H ( x- 1) = 1 Therefore, y = (1- x 2 )(1- 1) = 0 Sketch: 2 3. a) We need to solve f ( x ) = 0 in order to find the x-intercepts of f : f ( x ) = 0 = ⇒ x 3 + 3 x 2 + 3 x + 2 = 0 Notice that f (- 2) = 0 = ⇒ x + 2 is a factor of f ( x ). By long division, we see that f ( x ) = x 3 + 3 x 2 + 3 x + 2 = ( x + 2)( x 2 + x + 1) So f ( x ) = 0 = ⇒ x + 2 = 0 or x 2 + x + 1 = 0 x 2 + x + 1 = 0 = ⇒ x =- 1 ± p 1- 4(1)(1) 2 , which has no real solution. Therefore, the only x-intercept of f ( x ) is x =- 2. To find the vertical asymptotes of f ( x ), we set x 2- 1 = 0 = ⇒ x = ± 1 And since neither x- 1 nor x + 1 is a factor of the numerator of f , we find that f has vertical asymptotes at x = ± 1....
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This note was uploaded on 10/24/2009 for the course MATH 137 taught by Professor Speziale during the Fall '08 term at Waterloo.

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assignment1 solutions - MATH 137 Assignment #1 Solutions...

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