Hw1-solutions - meyer(akm995 Hw1 caputo(91795 This...

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meyer (akm995) – Hw1 – caputo – (91795) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points SimpliFy the di±erence quotient f ( x + h ) f ( x ) h , ( h n = 0) , as much as possible when f ( x ) = 5 x 2 + 4 x + 1 . 1. 10 x 4 + 5 h 2. 5 x 4 + 5 h 3. 10 x + 4 4. 5 x + 4 + 5 h 5. 10 x + 4 + 5 h correct Explanation: Now f ( x + h ) = 5( x + h ) 2 + 4( x + h ) + 1 = 5 x 2 + 10 xh + 4 x + 5 h 2 + 4 h + 1 , while f ( x ) = 5 x 2 + 4 x + 1 . Thus f ( x + h ) f ( x ) = 10 xh + 5 h 2 + 4 h . Consequently, f ( x + h ) f ( x ) h = 10 x + 4 + 5 h . 002 10.0 points Which Function has 1 2 3 4 5 6 1 2 3 4 as its graph on [ 1 , 7]? 1. f ( x ) = 1 1 x 2 + 1 | x 2 | 2. f ( x ) = 1 + 1 x 2 1 | x 2 | 3. f ( x ) = 1 + 1 x 2 + 1 | x 2 | correct 4. f ( x ) = 1 x 2 + 1 | x 2 | 1 5. f ( x ) = 1 1 x 2 1 | x 2 | Explanation: Since | x 2 | = b x 2 , x 2, 2 x, x < 2, we see that 1 x 2 + 1 | x 2 | = 2 x 2 , x > 2, 0 , x < 2 . Notice that the value x = 2 has to excluded because it does not belong to the natural domain oF 1 ( x 2) or oF 1 | x 2 | . In particular, the positive x -axis is a horizontal asymptote, while the line x = 2 For y > 0 is a vertical asymptote oF the graph 1 2 3 4 5 6 1 2 3 4
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meyer (akm995) – Hw1 – caputo – (91795) 2 of the function deFned by g ( x ) = 1 x 2 + 1 | x 2 | . The circle indicates that g ( x ) is not deFned at x = 2. Since the given graph is just the graph of g shifted vertically upwards by 1 units, the given graph is that of the function f ( x ) = 1 + 1 x 2 + 1 | x 2 | .
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Hw1-solutions - meyer(akm995 Hw1 caputo(91795 This...

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