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M408K Hm. 3 Solutions

# M408K Hm. 3 Solutions - Mahon Kevin Homework 3 Due 3:00 am...

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Mahon, Kevin – Homework 3 – Due: Sep 15 2005, 3:00 am – Inst: Edward Odell 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Suppose lim x 5 f ( x ) = 4 . Consider the following statements: A. Range of f need not contain 4. B. As f ( x ) approaches 4, x approaches 5. C. f is defined on ( a, b ) for some a < 5 < b . Which of these statements are true without further restrictions on f ? 1. C only 2. A only correct 3. all of them 4. B and C only 5. A and C only 6. B only 7. none of them 8. A and B only Explanation: A. True: ( f ( x ) need only APPROACH 4). B. Not True: ( f ( x ) approaches 4 AS x approaches 5). C. Not true: ( f ( x ) need not be defined at x = 5). keywords: Stewart5e, True/False, definition limit 002 (part 1 of 1) 10 points Below is the graph of a function f . -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 2 4 6 - 2 - 4 - 6 2 4 6 8 - 2 - 4 Use the graph to determine lim x 4 f ( x ) . 1. limit = 7 2. limit = 6 3. limit = 5 4. limit = 3 5. limit does not exist correct Explanation: From the graph it is clear the f has a left hand limit at x = 4 which is equal to 6; and a right hand limit which is equal to 2. Since the two numbers do not coincide, the limit does not exist . keywords: Stewart5e, limit, graph, limit at jump discontinuity 003 (part 1 of 1) 10 points Below is the graph of a function f .

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Mahon, Kevin – Homework 3 – Due: Sep 15 2005, 3:00 am – Inst: Edward Odell 2 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 2 4 - 2 - 4 2 4 - 2 - 4 Use the graph to determine lim x 2 f ( x ). 1. limit = - 1 2. limit = 1 3. does not exist 4. limit = 0 5. limit = - 2 correct Explanation: From the graph it is clear that the limit lim x 2 - f ( x ) = - 2 , from the left and the limit lim x 2+ f ( x ) = - 2 , from the right exist and coincide in value. Thus the two-sided lim x 2 f ( x ) = - 2 . keywords: Stewart5e, limit, graph, limit at removable discontinuity 004 (part 1 of 1) 10 points Consider the function f ( x ) = 3 - x, x < - 1 x, - 1 x < 3 ( x - 1) 2 , x 3 . Find all the values of a for which the limit lim x a f ( x ) exists, expressing your answer in interval no- tation. 1. ( -∞ , ) 2. ( -∞ , 3) (3 , ) 3. ( -∞ , - 1) ( - 1 , ) 4. ( -∞ , - 1] [3 , ) 5. ( -∞ , - 1) ( - 1 , 3) (3 , ) correct Explanation: The graph of f is a straight line on ( -∞ , - 1), so lim x a f ( x ) exists (and = f ( a )) for all a in ( -∞ , - 1). Similarly, the graph of f on ( - 1 , 3) is a straight line, so lim x a f ( x ) exists (and = f ( a )) for all a in ( - 1 , 3). On (3 , ), however, the graph of f is a parabola, so lim x a f ( x ) still exists (and = f ( a )) for all a in (3 , ). On the other hand, lim x → - 1 - f ( x ) = 4 , lim x → - 1+ f ( x ) = - 1 , while lim x 3 - f ( x ) = 3 , lim x 3+ f ( x ) = 4 . Thus neither of the limits lim x → - 1 f ( x ) , lim x 3 f ( x ) exists. Consequently, the limit exists only for values of a in ( -∞ , - 1) ( - 1 , 3) (3 , ) .
Mahon, Kevin – Homework 3 – Due: Sep 15 2005, 3:00 am – Inst: Edward Odell 3 keywords: Stewart5e, limit, linear function, piecewise-defined function, quadratic func- tion 005 (part 1 of 1) 10 points Let f be the function defined by f ( x ) = 12 x - 10 tan x sin 2 x .

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