Exam 3 - Garcia, Ilse – Exam 3 – Due: Dec 5 2007, 1:00...

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Unformatted text preview: Garcia, Ilse – Exam 3 – Due: Dec 5 2007, 1:00 am – Inst: Fonken 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 µ 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 Garcia, Ilse – Exam 3 – Due: Dec 5 2007, 1:00 am – Inst: Fonken 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 ≥ 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 note was uploaded on 02/22/2009 for the course M 58365 taught by Professor Gilbert during the Spring '08 term at University of Texas.

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Exam 3 - Garcia, Ilse – Exam 3 – Due: Dec 5 2007, 1:00...

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