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Unformatted text preview: padilla (tp5647) – Review Exam 02 – Fouli – (58320) 1 This printout should have 32 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points In the rightangled triangle θ 5 x the angle θ is increasing at a constant rate of 5 radians per hour. At what rate is the side of length x increas ing when x = 3 feet? 1. rate = 26 ft/hour 2. rate = 20 ft/hour correct 3. rate = 24 ft/hour 4. rate = 22 ft/hour 5. rate = 28 ft/hour Explanation: By rightangle trigonometry, sin θ = x 5 , so cos θ dθ dt = 1 5 dx dt , in which case dx dt = 5 cos θ dθ dt = 25 cos θ . But by the properties of 345 rightangle tri angles, cos θ = 4 / 5. Consequently, dx dt = 20 ft/hour . 002 10.0 points Which one of the following is the graph of f ( x ) = x x 2 4 ? 1. 2 4 2 4 2 4 2 4 2. 2 4 2 4 2 4 2 4 correct 3. 2 4 2 4 2 4 2 4 4. 2 4 2 4 2 4 2 4 padilla (tp5647) – Review Exam 02 – Fouli – (58320) 2 5. 2 4 2 4 2 4 2 4 Explanation: The graph will have vertical asymptotes at x = ± 2 and horizontal asymptote at y = 0. On the other hand, f ′ ( x ) = x 2 + 4 ( x 2 4) 2 , so f will be decreasing on (∞ , 2) , ( 2 , 2) , (2 , ∞ ) . The only graph having these is properties is 2 4 2 4 2 4 2 4 Consequently, this must be the graph of f . 003 10.0 points Determine the derivative of f ( x ) = x radicalbig x 2 1 . 1. f ′ ( x ) = 2 x 2 √ x 2 1 2. f ′ ( x ) = x 2 2 √ x 2 1 3. f ′ ( x ) = x 2 + 2 √ x 2 1 4. f ′ ( x ) = 1 2 x 2 √ x 2 1 5. f ′ ( x ) = 2 x 2 1 √ x 2 1 correct 6. f ′ ( x ) = 2 x 2 + 1 √ x 2 1 Explanation: By the Product and Chain Rules, f ′ ( x ) = radicalbig x 2 1 + x 2 √ x 2 1 = ( x 2 1) + x 2 √ x 2 1 . Consequently, f ′ ( x ) = 2 x 2 1 √ x 2 1 . keywords: 004 10.0 points The derivative, f ′ , of f has graph a b c graph of f ′ Use it to locate the critical point(s) x at which f has a local minimum? 1. x = b , c padilla (tp5647) – Review Exam 02 – Fouli – (58320) 3 2. x = a , b 3. x = b 4. x = c , a 5. x = a , b , c 6. x = c 7. x = a correct 8. none of a , b , c Explanation: Since the graph of f ′ ( x ) has no ‘holes’, the only critical points of f occur at the x intercepts of the graph of f ′ , i.e. , at x = a, b, and c . Now by the first derivative test, f will have (i) a local maximum at x if f ′ ( x ) changes from positive to negative as x passes through x ; (ii) a local minimum at x if f ′ ( x ) changes from negative to positive as x passes through x . Consequently, by looking at the sign of f ′ ( x ) near each of x = a, b, and c we see that f has a local minimum only at x = a ....
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This note was uploaded on 09/09/2009 for the course CH 302 taught by Professor Mccord during the Spring '09 term at University of TexasTyler.
 Spring '09
 MCCORD
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