Practice3 - Granillo Yvette – Review 3 – Due Dec 9 2005 6:00 pm – Inst Edward Odell 1 This print-out should have 18 questions Multiple-choice

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Unformatted text preview: Granillo, Yvette – Review 3 – Due: Dec 9 2005, 6:00 pm – Inst: Edward Odell 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 If f- 1 is the inverse of f , determine the value of f ( f- 1 (9)). 1. f ( f- 1 (9)) = 1 9 2. f ( f- 1 (9)) = 9 correct 3. f ( f- 1 (9)) = 81 4. Need to know f 5. f ( f- 1 (9)) = 1 81 Explanation: The inverse, f- 1 , of f has the property that f ( f- 1 ( x )) = x, f- 1 ( f ( x )) = x. Consequently, f ( f- 1 (9)) = 9 . keywords: Stewart5e, 002 (part 1 of 1) 10 points Find the inverse function, f- 1 , of f when f is defined by f ( x ) = √ 7 x- 5 , x ≥ 5 7 . 1. f- 1 ( x ) = 1 7 ( x 2 + 5) , x ≥ 7 5 2. f- 1 ( x ) = 1 5 p x 2 + 7 , x ≥ 7 5 3. f- 1 ( x ) = 1 7 ( x 2 + 5) , x ≥ correct 4. f- 1 ( x ) = 1 7 p x 2- 5 , x ≥ 5. f- 1 ( x ) = 1 5 ( x 2- 7) , x ≥ 5 7 6. f- 1 ( x ) = 1 5 p x 2- 7 , x ≥ Explanation: Since f has domain [ 5 7 , ∞ ) and is increasing on its domain, the inverse of f exists and has range [ 5 7 , ∞ ); furthermore, since f has range [0 , ∞ ), the inverse of f has domain [0 , ∞ ). To determine f- 1 we solve for x in y = √ 7 x- 5 and then interchange x, y . Solving first for x , we see that 7 x = y 2 + 5 . Consequently, f- 1 is defined on [0 , ∞ ) by f- 1 ( x ) = 1 7 ( x 2 + 5) . keywords: Stewart5e, 003 (part 1 of 1) 10 points Find the value of g (- 1) when g is the inverse of the function f defined by f ( x ) = 2 x 3- 3 x . ( Hint : f (1) =- 1.) 1. g (- 1) =- 2 2. g (- 1) = 1 9 correct 3. g (- 1) =- 1 9 4. g (- 1) = 2 5. g (- 1) =- 9 Granillo, Yvette – Review 3 – Due: Dec 9 2005, 6:00 pm – Inst: Edward Odell 2 6. g (- 1) = 9 Explanation: When g is the inverse of f then f ( g ( x )) = x, so f ( g ( x )) g ( x ) = 1 , i.e. , g ( x ) = 1 f ( g ( x )) . But, if f ( x ) = 2 x 3- 3 x , we know that f ( x ) = 6 x 2 + 3 x 2 , f (1) =- 1 ; in particular, g (- 1) = 1 and f (1) = 9 . Consequently, g (- 1) = 1 f (1) = 1 9 . keywords: Stewart5e, 004 (part 1 of 1) 10 points Find f ( x ) when f ( x ) = 9 cos x- 8 sin x and f (0) = 3. 1. f ( x ) = 9 sin x + 8 cos x- 6 2. f ( x ) = 9 cos x + 8 sin x- 6 3. f ( x ) = 9 cos x + 8 sin x + 12 4. f ( x ) =- 9 cos x + 8 sin x + 12 5. f ( x ) =- 9 sin x + 8 cos x- 5 6. f ( x ) = 9 sin x + 8 cos x- 5 correct Explanation: Since d dx sin x = cos x, d dx cos x =- sin x, we see immediately that f ( x ) = 9 sin x + 8 cos x + C with C an arbitrary constant. The condition f (0) = 3 determines C : f (0) = 3...
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This test prep was uploaded on 04/16/2008 for the course M 408k taught by Professor Schultz during the Spring '08 term at University of Texas at Austin.

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Practice3 - Granillo Yvette – Review 3 – Due Dec 9 2005 6:00 pm – Inst Edward Odell 1 This print-out should have 18 questions Multiple-choice

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