This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 xintercepts 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 xintercept 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....
View
Full
Document
This note was uploaded on 10/24/2009 for the course MATH 137 taught by Professor Speziale during the Fall '08 term at Waterloo.
 Fall '08
 SPEZIALE
 Math, Calculus

Click to edit the document details