HW1 (solutions)

HW1 (solutions) - myers(nmm698 HW01 Hamrick(54868 This...

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myers (nmm698) – HW01 – Hamrick – (54868) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points SimpliFy the expression f ( x ) = 3 + 9 x 2 1 3 p x x 2 4 P as much as possible. 1. f ( x ) = 3( x 2) x 4 2. f ( x ) = x 2 x + 4 3. f ( x ) = 3( x + 2) 2 x 4 4. f ( x ) = 3( x + 2) x 4 correct 5. f ( x ) = x + 2 x + 4 6. f ( x ) = x 2 2 x + 4 Explanation: AFter bringing the numerator to a common denominator it becomes 3 x 6 + 9 x 2 = 3 x + 3 x 2 . Similarly, aFter bringing the denominator to a common denominator and Factoring it be- comes x 2 4 3 x x 2 4 = ( x + 1)( x 4) x 2 4 . Consequently, f ( x ) = 3 + 9 x 2 1 3 p x x 2 4 P = 3 x + 3 ( x + 1)( x 4) p x 2 4 x 2 P . On the other hand, x 2 4 = ( x + 2)( x 2) . Thus, fnally, we see that f ( x ) = 3( x + 2) x 4 . 002 10.0 points Let f be the quadratic Function defned by f ( x ) = 2 x 2 12 x 14 . Use completing the square to fnd h so that f ( x ) = 2( x h ) 2 + k For some value oF k . 1. h = 3 correct 2. h = 6 3. h = 3 4. h = 6 5. h = 12 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 .
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myers (nmm698) – HW01 – Hamrick – (54868) 2 003 10.0 points Which, if any, of the following statements are true when a, b are real numbers? A. For all positive a and b , a + b = r a + 2 ab + b . B. For all a and b , r ( a + b ) 2 = a + b . C. For all positive a and b , a b a + b = a b . 1. A and C only correct 2. A and B only 3. B and C only 4. A only 5. C only 6. B only 7. all of them 8. none of them Explanation: A. TRUE: by the known product, ( x + y ) 2 = x 2 + 2 xy + y 2 . On the other hand, r ( x + y ) 2 = | x + y | , so if x + y > 0, x + y = R x 2 + 2 xy + y 2 . But if a, b are positive we can set x = a and y = b . The result follows since x and y are then positive. B. FALSE: r ( x + y ) 2 = | x + y | , and since R ( · ) is always non-negative, the right hand side has to be non-negative. But if a, b can be positive or negative, an absolute value sign is then needed on the right. C. TRUE: by the known di±erence 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 . The result follows after division. keywords: square root, properties of square root, PlaceUT, TrueFalse, T/F, 004 10.0 points By removing absolute values, express f ( x ) = 4 x 2 | x 2 | as a piecewise-de²ned function. 1. f ( x ) = b 2 x , x > 2 , x 2 , x < 2 . 2. f ( x ) = b x + 2 , x > 2 , ( x + 2) , x < 2 . 3. f ( x ) = b x + 2 , x > 2 , ( x + 2) , x < 2 . 4. f ( x ) = b ( x + 2) , x > 2 , x + 2 , x < 2 . cor- rect 5. f ( x ) = b x 2 , x > 2 , 2 x , x < 2 . 6. f ( x ) = b x 2 , x > 2 , 2 x , x < 2 .
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myers (nmm698) – HW01 – Hamrick – (54868) 3 Explanation: Since | a | = b a, , a 0 , a , a < 0 , we see that | x 2 | = b x 2 , x 2 , 2 x , x < 2 . In particular, therefore, x 2 | x 2 | = b 1 , x > 2 , 1 , x < 2 , where the point x = 2 has been excluded from the domain because the quotient is not deFned at that point. On the other hand, 4 x 2 = (2 + x )(2 x ) , so f can now be rewritten as a product f ( x ) = g ( x ) p x 2 | x 2 | P of g ( x ) = ( x +2) and the previous piecewise- deFned function.
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HW1 (solutions) - myers(nmm698 HW01 Hamrick(54868 This...

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