Calculus 408K Exam 3

# Calculus 408K Exam 3 - Barkley, Thane Exam 3 Due: Dec 5...

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Unformatted text preview: Barkley, Thane Exam 3 Due: Dec 5 2007, 1:00 am Inst: Louiza Fouli 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- . 28 2. Amount = \$300 e . 28 correct 3. Amount = \$300 e- 28 4. Amount = \$3 e 28 5. Amount = \$3 e . 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 . 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 where tan = x 5 ,- 2 < < 2 . To determine the value of sin given the value of tan , we can apply Pythagoras theorem to the right triangle Barkley, Thane Exam 3 Due: Dec 5 2007, 1:00 am Inst: Louiza Fouli 2 5 x p x 2 + 5 From this it follows that y = sin = x x 2 + 5 . Alternatively, we can use the trig identity csc 2 = 1 + cot 2 to determine sin . 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 2. f- 1 ( x ) = 1 4 ( x 2 + 7) , x 4 7 3. f- 1 ( x ) = 1 4 p x 2- 7 , x 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 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 (2)....
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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 at Austin.

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Calculus 408K Exam 3 - Barkley, Thane Exam 3 Due: Dec 5...

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