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Unformatted text preview: Barkley, Thane – Final 1 – Due: Dec 17 2007, 11:00 pm – Inst: Louiza Fouli 1 This printout should have 25 questions. Multiplechoice 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 twosided 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 µ 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 L’Hospital’s 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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This test prep was uploaded on 04/16/2008 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas.
 Fall '08
 schultz
 Calculus

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