HW01-solutions

HW01-solutions - white(taw933 – HW01 – ben-zvi...

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Unformatted text preview: white (taw933) – HW01 – ben-zvi – (55600) 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 10.0 points Simplify the expression f ( x ) = 3 + 15 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 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 + 15 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 + 15 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 . 002 10.0 points Let f be the quadratic function defined by f ( x ) = 2 x 2 + 12 x + 50 . Use completing the square to find h so that f ( x ) = 2( x + h ) 2 + k for some value of k . 1. h = 3 correct 2. h = − 6 3. h = 6 4. h = − 3 5. h = 12 Explanation: Completing the square gives f ( x ) = 2 x 2 + 12 x + 50 = 2( x 2 + 6 x + 25) = 2( x 2 + 6 x + 9 + 25 − 9) . Thus f ( x ) = 2( x 2 + 6 x + 9) + 32 = 2( x + 3) 2 + 32 . Consequently, h = 3 . white (taw933) – HW01 – ben-zvi – (55600) 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 = radicalBig a + 2 √ ab + b . B. For all positive a and b , a − b √ a + √ b = √ a − √ b . C. For all a and b , radicalBig ( a + b ) 2 = a + b . 1. B and C only 2. B only 3. C only 4. A and B only correct 5. A and C only 6. none of them 7. all of them 8. A only Explanation: A. TRUE: 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 . The result follows since x and y are then positive. B. TRUE: 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 . The result follows after division. C. FALSE: radicalBig ( x + y ) 2 = | x + y | , and since radicalbig ( · ) 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....
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This note was uploaded on 12/05/2011 for the course M 408k taught by Professor Schultz during the Spring '08 term at University of Texas.

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HW01-solutions - white(taw933 – HW01 – ben-zvi...

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