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# HW13 - Jolley Garrett Homework 13 Due 3:00 am Inst R...

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Jolley, Garrett – Homework 13 – Due: Nov 20 2007, 3:00 am – Inst: R Heitmann 1 This print-out should have 21 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 Find the solution of the exponential equa- tion 2 3 x = 4 x +12 . 1. x = - 24 2. x = 12 3. x = - 12 4. x = 24 correct 5. none of these Explanation: By properties of exponents, 4 x +12 = 2 2 x +24 . Thus the equation can be rewritten as 2 3 x = 2 2 x +24 , which after taking logs to the base 2 of both sides becomes 3 x = 2 x + 24 . Rearranging and solving we thus find that x = 24 . keywords: 002 (part 1 of 1) 10 points Which of the following is the graph of f ( x ) = 2 - x - 1 - 2 ? 1. 2 4 - 2 - 4 2 4 - 2 - 4 2. 2 4 - 2 - 4 2 4 - 2 - 4 3. 2 4 - 2 - 4 2 4 - 2 - 4 4. 2 4 - 2 - 4 2 4 - 2 - 4

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Jolley, Garrett – Homework 13 – Due: Nov 20 2007, 3:00 am – Inst: R Heitmann 2 5. 2 4 - 2 - 4 2 4 - 2 - 4 6. 2 4 - 2 - 4 2 4 - 2 - 4 correct Explanation: Since lim x → ∞ 2 - x = 0 , we see that lim x → ∞ f ( x ) = - 2 , in particular, f has a horizontal asymptote y = - 2. This eliminates all but two of the graphs. On the other hand, f (0) = - 3 2 , so the y -intercept of the given graph must occur at y = - 3 2 . Consequently, the graph is of f is 2 4 - 2 - 4 2 4 - 2 - 4 keywords: 003 (part 1 of 1) 10 points Find the value of lim x → ∞ 4 e 3 x + e - 3 x 2 e 3 x - 3 e - 3 x . 1. limit = - 3 5 2. limit = - 2 3. limit = - 1 2 4. limit = 3 5 5. limit = 2 correct 6. limit = 1 2 Explanation: After division we see that 4 e 3 x + e - 3 x 2 e 3 x - 3 e - 3 x = 4 + e - 6 x 2 - 3 e - 6 x . On the other hand, lim x → ∞ e - ax = 0 for all a > 0. But then by properties of limits, lim x → ∞ 4 + e - 6 x 2 - 3 e - 6 x = 2 . Consequently, limit = 2 . keywords: exponential function, limit as in- finity 004 (part 1 of 1) 10 points Find the value of f 0 ( - 2) when f ( x ) = x 2 + 4 e - 4 x .
Jolley, Garrett – Homework 13 – Due: Nov 20 2007, 3:00 am – Inst: R Heitmann 3 1. f 0 ( - 2) = - 4 - 16 e 8 correct 2. f 0 ( - 2) = - 4 + 4 e 8 3. f 0 ( - 2) = 8 + 4 e 8 4. f 0 ( - 2) = 8 - 16 e 8 5. f 0 ( - 2) = - 4 - 16 e 10 Explanation: By the Chain rule, df dx = 2 x - 16 e - 4 x Consequently, f 0 ( - 2) = - 4 - 16 e 8 . keywords: 005 (part 1 of 1) 10 points Determine the derivative of f ( x ) = e 2 x (cos 3 x + 5 sin 3 x ) . 1. f 0 ( x ) = e 2 x (7 cos 3 x + 17 sin 3 x ) 2. f 0 ( x ) = 2 e 2 x (3 cos 3 x + 15 sin 3 x ) 3. f 0 ( x ) = e 2 x (15 cos 3 x - 10 sin 3 x ) 4. f 0 ( x ) = e 2 x (17 cos 3 x + 7 sin 3 x ) cor- rect 5. f 0 ( x ) = e 2 x (17 cos 3 x - 7 sin 3 x ) 6. f 0 ( x ) = 2 e 2 x (15 cos 3 x - 3 sin 3 x ) Explanation: By the Product and Chain rules, f 0 ( x ) = 2 e 2 x (cos 3 x + 5 sin 3 x ) + e 2 x (15 cos 3 x - 3 sin 3 x ) .

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