Assignment 1 Solutions

# Assignment 1 Solutions - MATH 137 Fall 2010 Solutions to...

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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 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 | = 2 | x | 3 + | x | 2 + 3 | x | + | sin x | Since-1 sin x 1 for every real number x , then | sin x | 1. Since-2 x / 2 and / 2 &lt; 2, then | x | 2 for every x in this interval. Therefore, | 2 x 3-x 2 + 3 x-sin 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 &gt; When x &gt; 0, we have | x | = x . When x &gt; 0, we have-x &lt; 0 and so H ( x ) = 1 and H (-x ) = 0. Therefore, when x &gt; 0, we have xH ( x )-xH (-x ) = x 1-x 0 = x . Thus, | x | = xH ( x )-xH (-x ) in this case. Case 3: x &lt; When x &lt; 0, we have | x | =-x . When x &lt; 0, we have-x &gt; 0 and so H ( x ) = 0 and H (-x ) = 1. Therefore, when x &lt; 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 ) &lt; 0. Consider the given function y = H ( x 4 + x 3-7 x 2-x +6) and set f ( x ) = x 4 + x 3-7 x 2-x +6. To graph y = H ( f ( x )), we need to determine where f ( x ) 0 and where f ( x ) &lt; 0....
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## 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 .

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Assignment 1 Solutions - MATH 137 Fall 2010 Solutions to...

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