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Unformatted text preview: myers (nmm698) – HW10 – Hamrick – (54868) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Let f be the function defined by f ( x ) = 1 + x 2 / 3 . Consider the following properties: A. has local maximum at x = 0 B. concave down on ( −∞ , 0) ∪ (0 , ∞ ) Which does f have? 1. A only 2. B only correct 3. neither of them 4. both of them Explanation: The graph of f is 2 4 − 2 − 4 2 4 On the other hand, after differentiation, f ′ ( x ) = 2 3 x 1 / 3 , f ′′ ( x ) = − 2 9 x 4 / 3 . Consequently, A. FALSE: see graph. B. TRUE: f ′′ ( x ) < , x negationslash = 0 002 10.0 points The graph of a twicedifferentiable function f is shown in 1 Which one of the following sets of inequalities is satisfied by f and its derivatives at x = 1? 1. f ′ (1) < f ′′ (1) < f (1) 2. f (1) < f ′′ (1) < f ′ (1) 3. f ′′ (1) < f ′ (1) < f (1) correct 4. f ′′ (1) < f (1) < f ′ (1) 5. f (1) < f ′ (1) < f ′′ (1) 6. f ′ (1) < f (1) < f ′′ (1) Explanation: (i) Whether the point (1 , f (1)) lies above or below the xaxis determines the sign of f (1), and f (1) = 0 if the point lies on the xaxis. (ii) Whether the graph of f is increasing or decreasing at the point (1 , f (1)) deter mines the sign of f ′ (1), and f ′ (1) = 0 when the tangent line to the graph is horizontal at (1 , f (1)). (iii) Whether the graph of f is concave up or concave down at the point (1 , f (1)) determines the sign of f ′′ (1), and if the graph is changing concavity at (1 , f (1)), then f ′′ (1) = 0. In the graph above, therefore, the inequali ties f ′′ (1) < f ′ (1) < f (1) are satisfied. myers (nmm698) – HW10 – Hamrick – (54868) 2 003 10.0 points When Sue uses first and second derivatives to analyze a particular continuous function y = f ( x ) she obtains the chart y y ′ y ′′ x < − 3 + − x = − 3 4 − 3 < x < − − x = 0 1 − 1 < x < 2 − + x = 2 − 1 DNE x > 2 + + Which of the following can she conclude from her chart? A. f is concave down on ( −∞ , 0) . B. f has a point of inflection at x = 0. C. f is concave up on (0 , 2) . 1. none of them 2. A and C only 3. all of them correct 4. B and C only 5. B only 6. A and B only 7. C only 8. A only Explanation: The graph of f must look like 2 − 2 − 4 2 4 Consequently, A. True. B. True. C. True. 004 10.0 points Find the interval(s) where f ( x ) = x 4 − 5 x 3 − 36 x 2 − 5 x + 2 is concave down. 1. parenleftBig −∞ , − 3 2 parenrightBig , parenleftBig 4 , ∞ parenrightBig 2. parenleftBig −∞ , − 4 parenrightBig , parenleftBig − 3 2 , ∞ parenrightBig 3....
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This note was uploaded on 03/10/2011 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas.
 Fall '08
 schultz

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