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Unformatted text preview: Create assignment, 57321, Homework 5, Feb 22 at 4:21 pm 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. CalC4a03a 50:01, calculus3, multiple choice, < 1 min, wordingvariable. 001 If f is the function whose graph is given by 2 4 6 2 4 6 which of the following properties does f NOT have? 1. lim x → 4 f ( x ) = 5 correct 2. local maximum at x = 4 3. critical point at x = 2 4. lim x → 2 + f ( x ) = lim x → 2 f ( x ) 5. f ( x ) > 0 on (4 , 7) Explanation: The given graph has a removable disconti nuity at x = 4. On the other hand, recall that f has a local maximum at a point c when f ( x ) ≤ f ( c ) for all x near c . Thus f could have a local maximum even if the graph of f has a removable discontinuity at c ; simi larly, the definition of local minimum allows the graph of f to have a local minimum at a removable disconituity. So it makes sense to ask if f has a local extremum at x = 4. Inspection of the graph now shows that the only property f does not have is lim x → 4 f ( x ) = 5 . keywords: graph, limit, removable discontinu ity critical point, left hand limit, right hand limit, slope CalC4a08s 50:01, calculus3, multiple choice, > 1 min, fixed. 002 If f is a continuous function on [1 , 5] having (1) an absolute minimum at 1, (2) an absolute maximum at 5, (3) a local maximum at 2, and (4) a local minimum at 4, which of the following could be the graph of such a function f ? 1. 1 2 3 4 5 1 2 3 4 5 x y cor rect 2. 1 2 3 4 5 1 2 3 4 5 x y Create assignment, 57321, Homework 5, Feb 22 at 4:21 pm 2 3. 1 2 3 4 5 1 2 3 4 5 x y 4. 1 2 3 4 5 1 2 3 4 5 x y Explanation: keywords: absolute minimum, absolute maxi mum, local maximum, local minimum CalC4a37a 50:01, calculus3, multiple choice, < 1 min, wordingvariable. 003 Find all the critical points of the function f ( x ) = 2sin x x  on the interval ( π, π ). 1. x = 2 π 3 , , π 3 correct 2. x = π 3 , , 2 π 3 3. x = 0 4. x = π 6 , , π 6 5. x = π 6 , π 6 6. x = 5 π 6 , 5 π 6 7. x = π 3 , 2 π 3 8. x = 2 π 3 , π 3 Explanation: Since  x  is differentiable everywhere except at x = 0, while sin x is differentiable for all x , the point x = 0 is a critical point of f and all other critical points will be the solutions of f ( x ) = 0. Now f ( x ) = ‰ 2cos x 1 , x > 0, 2cos x + 1 , x < 0. But on (0 , π ) cos x = 1 2 = ⇒ x = π 3 , while on ( π, 0) cos x = 1 2 = ⇒ x = 2 π 3 . Consequently, the only critical points of f on ( π, π ) are x = 2 π 3 , , π 3 ....
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This note was uploaded on 03/20/2008 for the course M 408c taught by Professor Mcadam during the Spring '06 term at University of Texas.
 Spring '06
 McAdam

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