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HW1-solutions - wei(jw35975 HW1 milburn(54685 This...

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wei (jw35975) – HW1 – milburn – (54685) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Which, if any, of the following statements are true when a, b are real numbers? A. For all positive a and b . a b a + b = a + b . B. For all a and b , radicalBig ( a + b ) 2 = | a + b | . C. For all positive a and b , a + b = a + b . 1. B and C only 2. none of them 3. B only correct 4. A only 5. all of them 6. A and C only 7. A and B only 8. C only Explanation: A. FALSE: by the known difference of squares factorization, x 2 y 2 = ( x y )( x + y ) . But if a, b are positive we can set x = a and y = b . Thus, after division, a b a + b = a b , contrary to the assertion. B. TRUE: we know that radicalBig ( x + y ) 2 = | x + y | , and since radicalbig ( · ) is always non-negative, the right hand side has to be non-negative. That’s why the absolute value sign is needed. C. FALSE: by the known product, ( x + y ) 2 = x 2 + 2 xy + y 2 . On the other hand, radicalBig ( x + y ) 2 = | x + y | , so if x + y > 0, x + y = radicalbig x 2 + 2 xy + y 2 . But if a, b are positive we can set x = a and y = b , in which case a + b = radicalBig a + 2 ab + b , contrary to the assertion. keywords: square root, properties of square root, PlaceUT, TrueFalse, T/F, 002 10.0points Simplify the expression f ( x ) = 3 + 9 x 4 1 + 15 parenleftBig x x 2 16 parenrightBig as much as possible. 1. f ( x ) = 3( x 4) x + 16 2. f ( x ) = x + 4 x 16 3. f ( x ) = 3( x + 4) 2 x + 16
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wei (jw35975) – HW1 – milburn – (54685) 2 4. f ( x ) = 3( x + 4) x + 16 correct 5. f ( x ) = x 4 x 16 6. f ( x ) = x 4 2 x 16 Explanation: After bringing the numerator to a common denominator it becomes 3 x 12 + 9 x 4 = 3 x 3 x 4 . Similarly, after bringing the denominator to a common denominator and factoring it be- comes x 2 16 + 15 x x 2 16 = ( x 1)( x + 16) x 2 16 . Consequently, f ( x ) = 3 + 9 x 4 1 + 15 parenleftBig x x 2 16 parenrightBig = 3 x 3 ( x 1)( x + 16) parenleftBig x 2 16 x 4 parenrightBig . On the other hand, x 2 16 = ( x + 4)( x 4) . Thus, finally, we see that f ( x ) = 3( x + 4) x + 16 . 003 10.0points Let f be the quadratic function defined by f ( x ) = 2 x 2 12 x 14 . Use completing the square to find h so that f ( x ) = 2( x h ) 2 + k for some value of k . 1. h = 12 2. h = 6 3. h = 3 correct 4. h = 3 5. h = 6 Explanation: Completing the square gives f ( x ) = 2 x 2 12 x 14 = 2( x 2 6 x 7) = 2( x 2 6 x + 9 7 9) . Thus f ( x ) = 2( x 2 6 x + 9) 32 = 2( x 3) 2 32 . Consequently, h = 3 . 004 10.0points By removing absolute values, express f ( x ) = 4 x 2 | x + 2 | as a piecewise-defined function. 1. f ( x ) = braceleftbigg x 2 , x > 2 , 2 x , x < 2 . 2. f ( x ) = braceleftbigg x 2 , x > 2 , 2 x , x < 2 . 3. f ( x ) = braceleftbigg 2 x , x > 2 , x 2 , x < 2 . cor- rect 4. f ( x ) = braceleftbigg x + 2 , x > 2 , ( x + 2) , x < 2 .
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wei (jw35975) – HW1 – milburn – (54685) 3 5. f ( x ) = braceleftbigg ( x + 2) , x > 2 , x + 2 , x < 2 .
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