Homework 13 - Garcia, Ilse Homework 13 Due: Nov 20 2007,...

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Garcia, Ilse – Homework 13 – Due: Nov 20 2007, 3:00 am – Inst: Fonken 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. The due time is Central time. 001 (part 1 o± 1) 10 points Find the solution o± the exponential equa- tion 2 8 x = 4 3 2 x +9 . 1. x = 3 2. x = - 3 3. x = - 18 5 4. x = 18 5 correct 5. none o± these Explanation: By properties o± exponents, 4 3 2 x +9 = 2 3 x +18 . Thus the equation can be rewritten as 2 8 x = 2 3 x +18 , which a±ter taking logs to the base 2 o± both sides becomes 8 x = 3 x + 18 . Rearranging and solving we thus fnd that x = 18 5 . keywords: 002 (part 1 o± 1) 10 points Which o± the ±ollowing is the graph o± f ( x ) = 2 - 3 - x ? 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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Garcia, Ilse – Homework 13 – Due: Nov 20 2007, 3:00 am – Inst: Fonken 2 5. 2 4 - 2 - 4 2 4 - 2 - 4 6. 2 4 - 2 - 4 2 4 - 2 - 4 correct Explanation: Since lim x →∞ 3 - 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) = 1, so the y -intercept of the given graph must occur at y = 1. 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 →∞ µ 3 e 3 x + 5 e - 3 x 4 e 3 x - e - 3 x . 1. limit = 4 3 2. limit = 2 5 3. limit = - 4 3 4. limit = 3 4 correct 5. limit = - 3 4 6. limit = - 2 5 Explanation: After division we see that 3 e 3 x + 5 e - 3 x 4 e 3 x - e - 3 x = 3 + 5 e - 6 x 4 - e - 6 x . On the other hand, lim x →∞ e - ax = 0 for all a > 0. But then by properties of limits, lim x →∞ 3 + 5 e - 6 x 4 - e - 6 x = 3 4 . Consequently, limit = 3 4 . keywords: exponential function, limit as in- ±nity 004 (part 1 of 1) 10 points Find the value of f 0 ( - 1) when f ( x ) = x 2 + e - 3 x .
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Garcia, Ilse – Homework 13 – Due: Nov 20 2007, 3:00 am – Inst: Fonken 3 1. f 0 ( - 1) = - 2 - 3 e 3 correct 2. f 0 ( - 1) = 2 - 3 e 3 3. f 0 ( - 1) = 2 + e 3 4. f 0 ( - 1) = - 2 + e 3 5. f 0 ( - 1) = - 2 - 3 e 4 Explanation: By the Chain rule, df dx = 2 x - 3 e - 3 x Consequently, f 0 ( - 1) = - 2 - 3 e 3 . keywords: 005 (part 1 of 1) 10 points Determine the derivative of f ( x ) = e 5 x (cos 2 x + 5 sin 2 x ) . 1. f 0 ( x ) = e 5 x (15 cos 2 x - 23 sin 2 x ) 2. f 0 ( x ) = e 5 x (15 cos 2 x + 23 sin 2 x ) cor- rect 3. f 0 ( x ) = 5 e 5 x (10 cos 2 x - 2 sin 2 x ) 4. f 0 ( x ) = e 5 x (10 cos 2 x - 25 sin 2 x ) 5. f 0 ( x ) = e 5 x (23 cos 2 x + 15 sin 2 x ) 6. f 0 ( x ) = 5 e 5 x (2 cos 2 x + 10 sin 2 x ) Explanation: By the Product and Chain rules, f 0 ( x ) = 5 e 5 x (cos 2 x + 5 sin 2 x ) + e 5 x (10 cos 2 x - 2 sin 2 x ) . Consequently,
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Homework 13 - Garcia, Ilse Homework 13 Due: Nov 20 2007,...

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