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# Exam 3 - Jolley Garrett – Exam 3 – Due Dec 5 2007 1:00...

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Unformatted text preview: Jolley, Garrett – Exam 3 – Due: Dec 5 2007, 1:00 am – Inst: R Heitmann 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 = 8 log 3 ‡ 1 9 · + 9 log 4 16 . 1. x = 2 correct 2. x =- 4 3. x = 4 4. x =- 3 5. x =- 2 6. x = 3 Explanation: By properties of logs, log 3 ‡ 1 9 · = log 3 ‡ 1 (3) 2 · =- 2 , while log 4 16 = log 4 (4) 2 = 2 . Consequently, x = 18- 16 = 2 . keywords: 002 (part 1 of 1) 10 points If \$200 is invested at an annual interest rate of 8%, determine the value of the investment after 4 years when interest is compounded continuously, leaving your answer in expo- nential form. 1. Amount = \$2 e . 32 2. Amount = \$200 e- . 32 3. Amount = \$200 e- 32 4. Amount = \$200 e . 32 correct 5. Amount = \$2 e 32 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 = 200, r = 8 and n = 4, therefore, Amount = \$200 e . 32 . keywords: 003 (part 1 of 1) 10 points Simplify the expression y = sin µ tan- 1 x √ 6 ¶ by writing it in algebraic form. 1. y = x √ x 2 + 6 correct 2. y = √ 6 √ x 2 + 6 3. y = √ x 2 + 6 √ 6 4. y = x √ x 2- 6 5. y = x x 2 + 6 Explanation: The given expression has the form y = sin θ where tan θ = x √ 6 ,- 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 Jolley, Garrett – Exam 3 – Due: Dec 5 2007, 1:00 am – Inst: R Heitmann 2 √ 6 x θ p x 2 + 6 From this it follows that y = sin θ = x √ x 2 + 6 . 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 ) = √ 5 x- 6 , x ≥ 6 5 . 1. f- 1 ( x ) = 1 5 p x 2- 6 , x ≥ 2. f- 1 ( x ) = 1 5 ( x 2 + 6) , x ≥ 5 6 3. f- 1 ( x ) = 1 6 p x 2- 5 , x ≥ 4. f- 1 ( x ) = 1 5 ( x 2 + 6) , x ≥ correct 5. f- 1 ( x ) = 1 6 ( x 2- 5) , x ≥ 6 5 6. f- 1 ( x ) = 1 6 p x 2 + 5 , x ≥ 5 6 Explanation: Since f has domain [ 6 5 , ∞ ) and is increasing on its domain, the inverse of f exists and has range [ 6 5 , ∞ ); furthermore, since f has range [0 , ∞ ), the inverse of f has domain [0 , ∞ ). To determine f- 1 we solve for x in y = √ 5 x- 6 and then interchange x, y . Solving first for x , we see that 5 x = y 2 + 6 . Consequently, f- 1 is defined on [0 , ∞ ) by f- 1 ( x ) = 1 5 ( x 2 + 6) . keywords: 005 (part 1 of 1) 10 points When g is the inverse of f ( x ) = x 3 + 2 x- 1 , find the value of g (11)....
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Exam 3 - Jolley Garrett – Exam 3 – Due Dec 5 2007 1:00...

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