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# EXAM3 - Vu Khanh Exam 3 Due Dec 5 2007 1:00 am Inst Samuels...

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Vu, Khanh – Exam 3 – Due: Dec 5 2007, 1:00 am – Inst: Samuels 1 This print-out should have 18 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 value of x when x = 7 log 2 1 4 · + 8 log 3 9 . 1. x = - 3 2. x = 4 3. x = 2 correct 4. x = - 2 5. x = - 4 6. x = 3 Explanation: By properties of logs, log 2 1 4 · = log 2 1 (2) 2 · = - 2 , while log 3 9 = log 3 (3) 2 = 2 . Consequently, x = 16 - 14 = 2 . keywords: 002 (part 1 of 1) 10 points If \$300 is invested at an annual interest rate of 7%, determine the value of the investment after 4 years when interest is compounded continuously, leaving your answer in expo- nential form. 1. Amount = \$300 e - 0 . 28 2. Amount = \$300 e 0 . 28 correct 3. Amount = \$300 e - 28 4. Amount = \$3 e 28 5. Amount = \$3 e 0 . 28 Explanation: When \$ P is invested at an annual interest rate of r % compounded continuously, then af- ter n years the investment is worth \$ Pe rn/ 100 . When P = 300, r = 7 and n = 4, therefore, Amount = \$300 e 0 . 28 . keywords: 003 (part 1 of 1) 10 points Simplify the expression y = sin tan - 1 x 5 by writing it in algebraic form. 1. y = 5 x 2 + 5 2. y = x 2 + 5 5 3. y = x x 2 + 5 4. y = x x 2 + 5 correct 5. y = x x 2 - 5 Explanation: The given expression has the form y = sin θ where tan θ = x 5 , - π 2 < θ < π 2 . To determine the value of sin θ given the value of tan θ , we can apply Pythagoras’ theorem to the right triangle

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Vu, Khanh – Exam 3 – Due: Dec 5 2007, 1:00 am – Inst: Samuels 2 5 x θ p x 2 + 5 From this it follows that y = sin θ = x x 2 + 5 . Alternatively, we can use the trig identity csc 2 θ = 1 + cot 2 θ to determine sin θ . keywords: 004 (part 1 of 1) 10 points Find the inverse function, f - 1 , of f when f is defined by f ( x ) = 4 x - 7 , x 7 4 . 1. f - 1 ( x ) = 1 7 p x 2 - 4 , x 0 2. f - 1 ( x ) = 1 4 ( x 2 + 7) , x 4 7 3. f - 1 ( x ) = 1 4 p x 2 - 7 , x 0 4. f - 1 ( x ) = 1 7 p x 2 + 4 , x 4 7 5. f - 1 ( x ) = 1 7 ( x 2 - 4) , x 7 4 6. f - 1 ( x ) = 1 4 ( x 2 + 7) , x 0 correct Explanation: Since f has domain [ 7 4 , ) and is increasing on its domain, the inverse of f exists and has range [ 7 4 , ); furthermore, since f has range [0 , ), the inverse of f has domain [0 , ). To determine f - 1 we solve for x in y = 4 x - 7 and then interchange x, y . Solving first for x , we see that 4 x = y 2 + 7 . Consequently, f - 1 is defined on [0 , ) by f - 1 ( x ) = 1 4 ( x 2 + 7) . keywords: 005 (part 1 of 1) 10 points When g is the inverse of f ( x ) = x 3 + 3 x - 2 , find the value of g 0 (2). ( Hint : f (1) = 2.) 1. g 0 (2) = 1 3 2. g 0 (2) = 2 3 3. g 0 (2) = 6 4. g 0 (2) = 3 2 5. g 0 (2) = 3 6. g 0 (2) = 1 6 correct Explanation: By definition, g ( f ( x )) = x , so by the Chain Rule, g 0 ( f ( x )) f 0 ( x ) = 1 . Thus g 0 ( f ( x )) = 1 f 0 ( x ) .
Vu, Khanh – Exam 3 – Due: Dec 5 2007, 1:00 am – Inst: Samuels 3 To determine the value of g 0 ( a ), therefore, we need to know b so that f ( b ) = a , for then g 0 ( a ) = 1 f 0 ( b ) . In the given example, f ( x ) = x 3 + 3 x - 2 , a = 2 . Then f 0 ( x ) = 3 x 2 + 3 , f (1) = 2 , in particular, b = 1. Consequently, g 0 (2) = 1 f 0 (1) = 1 6 .

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EXAM3 - Vu Khanh Exam 3 Due Dec 5 2007 1:00 am Inst Samuels...

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