Solutions - Exam 2

Solutions - Exam 2 - MATHEMATICS 111 - EXAM II GREG...

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MATHEMATICS 111 - EXAM II GREG RAINWATER WINTER 2010 Name: SOLUTIONS Date: 02/19/10 Question Pts. Worth Score #1 10 #2 10 #3 10 #4 25 #5 10 #6 10 #7 10 #8 15 Total 100 1
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2 EXAM II - MATH 111 #1 (a) What is the equation of the curve y = f ( x ) = x 3 - 1 x after transforming it 1 unit down and 2 units right ? Solution: (i) Translate y = f ( x ) down 1 unit: y = f ( x ) + 1 = x 3 - 1 x - 1 (ii) Translate y = f ( x ) - 1 right 2 units: y = f ( x - 2) - 1 = ( x - 2) 3 - 1 x - 2 - 1 y = ( x - 2) 3 - 1 x - 2 - 1 (b) What is the equation of the curve y = f ( x ) = x 2 - x after transforming it 1 unit up and reflected about the y-axis . Solution: (i) Reflect y = f ( x ) about the y-axis: y = f ( - x ) = ( - x ) 2 - ( - x ) = x 2 + x (ii) Translate y = f ( - x ) up 1 unit: y = f ( - x ) + 1 = x 2 + x + 1. y = x 2 + x + 1
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EXAM II - MATH 111 3 #2 Determine if f is even, odd, or neither. (a) f ( x ) = x 2 - x 3 Solution: f ( - x ) = ( - x ) 2 - ( - x ) 3 = x 2 + x 3 , - f ( x ) = - ( x 2 - x 3 ) = - x 2 + x 3 Since f ( - x ) 6 = f ( x ), f is not even. Since f ( - x ) 6 = - f ( x ), f is not odd. Therefore f must be neither odd or even. (b) f ( x ) = 1 1 + x 2 Solution: f ( - x ) = 1 1 + ( - x ) 2 = 1 1 + x 2 = f ( x ) Since f ( - x ) = f ( x ) for all x in the domain of f , f
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This note was uploaded on 09/18/2011 for the course WR 227 taught by Professor U.nknown during the Spring '10 term at Portland State.

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Solutions - Exam 2 - MATHEMATICS 111 - EXAM II GREG...

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