Calculus 408K Final - Barkley, Thane Final 1 Due: Dec 17...

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Unformatted text preview: Barkley, Thane Final 1 Due: Dec 17 2007, 11:00 pm Inst: Louiza Fouli 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 At which point on the graph A B C D E F is the slope greatest ( i.e. , most positive)? 1. B 2. C correct 3. A 4. E 5. D 6. F Explanation: By inspection the point is C . keywords: slope, graph, change of slope 002 (part 1 of 1) 10 points Below is the graph of a function f . 2 4- 2- 4 2 4- 2- 4 Use the graph to determine lim x 3 f ( x ). 1. does not exist 2. limit =- 1 3. limit =- 3 correct 4. limit = 0 5. limit =- 2 Explanation: From the graph it is clear that the limit lim x 3- f ( x ) =- 3 , from the left and the limit lim x 3+ f ( x ) =- 3 , from the right exist and coincide in value. Thus the two-sided lim x 3 f ( x ) =- 3 . keywords: limit, graph, limit at removable discontinuity 003 (part 1 of 1) 10 points Determine the limit lim x 6 8 (6- x ) 2 . 1. none of the other answers Barkley, Thane Final 1 Due: Dec 17 2007, 11:00 pm Inst: Louiza Fouli 2 2. limit = correct 3. limit = 4 3 4. limit =- 4 3 5. limit =- Explanation: Since lim x ( x- 6) 2 = 0 and (6- x ) 2 > 0 for all x 6 = 0, we see that lim x 6 8 (6- x ) 2 = . keywords: limit, rational function 004 (part 1 of 1) 10 points Determine if lim x x p x 2 + 4- x exists, and if it does, find its value. 1. limit = 1 2. limit = 3 3. limit = 2 correct 4. limit = 5 2 5. limit does not exist 6. limit = 3 2 Explanation: After rationalization we see that x p x 2 + 4- x = x x 2 + 4- x 2 x 2 + 4 + x = 4 x x 2 + 4 + x = 4 r 1 + 4 x 2 + 1 . On the other hand, lim x r 1 + 4 x 2 = 1 . Consequently, lim x x p x 2 + 4- x exists and has limit = 2 . keywords: limit, limit at infinity, square root function, rationalize numerator 005 (part 1 of 1) 10 points Find the value of lim x e 3 x- e- 3 x sin 2 x . 1. limit = 7 2 2. limit does not exist 3. limit = 3 2 4. limit = 3 correct 5. limit = 4 6. limit = 2 Explanation: Set f ( x ) = e 3 x- e- 3 x , g ( x ) = sin 2 x. Then f, g are everywhere differentiable func- tions such that lim x f ( x ) = lim x g ( x ) = 0 . Thus LHospitals Rule applies: lim x f ( x ) g ( x ) = lim x f ( x ) g ( x ) . Barkley, Thane Final 1 Due: Dec 17 2007, 11:00 pm Inst: Louiza Fouli 3 Now f ( x ) = 3( e 3 x + e- 3 x ) , g ( x ) = 2 cos 2 x, while lim x f ( x ) = 6 , lim x g ( x ) = 2 . Consequently, lim x e 3 x- e- 3 x sin 2 x = 3 . keywords: 006 (part 1 of 1) 10 points Find all the values of x at which the func- tion f defined by f ( x ) = x- 4 2 x 2- 4 x- 16 is not continuous....
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Calculus 408K Final - Barkley, Thane Final 1 Due: Dec 17...

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