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Unformatted text preview: Tran, Tony – Exam 3 – Due: Dec 5 2007, 1:00 am – Inst: Samuels 1 This printout should have 18 questions. Multiplechoice 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 = 6 log 2 ‡ 1 4 · + 7 log 3 27 . 1. x = 11 2. x = 10 3. x = 9 4. x = 9 correct 5. x = 10 6. x = 11 Explanation: By properties of logs, log 2 ‡ 1 4 · = log 2 ‡ 1 (2) 2 · = 2 , while log 3 27 = log 3 (3) 3 = 3 . Consequently, x = 21 12 = 9 . keywords: 002 (part 1 of 1) 10 points If $600 is invested at an annual interest rate of 8%, determine the value of the investment after 5 years when interest is compounded continuously, leaving your answer in expo nential form. 1. Amount = $600 e . 4 2. Amount = $600 e . 4 correct 3. Amount = $6 e . 4 4. Amount = $6 e 40 5. Amount = $600 e 40 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 = 600, r = 8 and n = 5, therefore, Amount = $600 e . 4 . keywords: 003 (part 1 of 1) 10 points Simplify the expression y = sin µ tan 1 x √ 11 ¶ by writing it in algebraic form. 1. y = x √ x 2 + 11 correct 2. y = √ 11 √ x 2 + 11 3. y = x x 2 + 11 4. y = √ x 2 + 11 √ 11 5. y = x √ x 2 11 Explanation: The given expression has the form y = sin θ where tan θ = x √ 11 , 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 Tran, Tony – Exam 3 – Due: Dec 5 2007, 1:00 am – Inst: Samuels 2 √ 11 x θ p x 2 + 1 1 From this it follows that y = sin θ = x √ x 2 + 11 . 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 ) = √ 3 x 5 , x ≥ 5 3 . 1. f 1 ( x ) = 1 5 p x 2 + 3 , x ≥ 3 5 2. f 1 ( x ) = 1 3 ( x 2 + 5) , x ≥ 3 5 3. f 1 ( x ) = 1 3 p x 2 5 , x ≥ 4. f 1 ( x ) = 1 5 ( x 2 3) , x ≥ 5 3 5. f 1 ( x ) = 1 5 p x 2 3 , x ≥ 6. f 1 ( x ) = 1 3 ( x 2 + 5) , x ≥ correct Explanation: Since f has domain [ 5 3 , ∞ ) and is increasing on its domain, the inverse of f exists and has range [ 5 3 , ∞ ); furthermore, since f has range [0 , ∞ ), the inverse of f has domain [0 , ∞ ). To determine f 1 we solve for x in y = √ 3 x 5 and then interchange x, y . Solving first for x , we see that 3 x = y 2 + 5 . Consequently, f 1 is defined on [0 , ∞ ) by f 1 ( x ) = 1 3 ( x 2 + 5) . keywords: 005 (part 1 of 1) 10 points When g is the inverse of f ( x ) = x 3 + 5 x 4 , find the value of g (14)....
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 Fall '08
 schultz
 Calculus, Derivative, Sin, Limit of a function, Exponentiation, Indeterminate form

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