This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MATH 137 Fall 2010 Solutions to Assignment #1, Part II 1. By the Triangle Inequality and properties of the absolute value,  2 x 3x 2 + 3 xsin x  =  2 x 3 + (x 2 ) + 3 x + (sin x )   2 x 3  +  x 2  +  3 x  +  sin x  = 2  x 3  +  x 2  + 3  x  +  sin x  = 2  x  3 +  x  2 + 3  x  +  sin x  Since1 sin x 1 for every real number x , then  sin x  1. Since2 x / 2 and / 2 < 2, then  x  2 for every x in this interval. Therefore,  2 x 3x 2 + 3 xsin x  2  x  3 +  x  2 + 3  x  +  sin x  = 2(2 3 ) + 2 2 + 3(2) + 1 = 27 as required. 2. (a) We consider three cases separately. Case 1: x = 0 When x = 0, we have  x  =   = 0. When x = 0, we have xH ( x )xH (x ) = 0 H (0)H (0) = 0. Therefore,  x  = xH ( x )xH (x ) in this case. Case 2: x > When x > 0, we have  x  = x . When x > 0, we havex < 0 and so H ( x ) = 1 and H (x ) = 0. Therefore, when x > 0, we have xH ( x )xH (x ) = x 1x 0 = x . Thus,  x  = xH ( x )xH (x ) in this case. Case 3: x < When x < 0, we have  x  =x . When x < 0, we havex > 0 and so H ( x ) = 0 and H (x ) = 1. Therefore, when x < 0, we have xH ( x )xH (x ) = x x 1 =x . Thus,  x  = xH ( x )xH (x ) in this case. Therefore, for every real number x , we have  x  = xH ( x )xH (x ). (b) When we have a function of the form y = H ( f ( x )), then by the denition of the Heaviside function, we will have y = 1 whenever f ( x ) 0 and y = 0 whenever f ( x ) < 0. Consider the given function y = H ( x 4 + x 37 x 2x +6) and set f ( x ) = x 4 + x 37 x 2x +6. To graph y = H ( f ( x )), we need to determine where f ( x ) 0 and where f ( x ) < 0....
View
Full
Document
This note was uploaded on 12/18/2010 for the course ECONOMICS 120 taught by Professor Mesta during the Spring '10 term at Wilfred Laurier University .
 Spring '10
 Mesta

Click to edit the document details