K Exam 2-solutions-1

# K Exam 2-solutions-1 - Version 020 – K Exam 2 – ben-zvi...

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Unformatted text preview: Version 020 – K Exam 2 – ben-zvi – (55600) 1 This print-out should have 18 questions. Multiple-choice 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 ) = 5 − x 2 / 3 . Consider the following properties: A. concave up on ( −∞ , 0) ∪ (0 , ∞ ) B. has local maximum at x = 0 Which does f have? 1. both of them correct 2. B only 3. A only 4. neither 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. TRUE: f ′′ ( x ) > , x negationslash = 0 B. TRUE: see graph 002 10.0 points Determine the interval(s) where f ( x ) = x − sin x is increasing on [0 , 2 π ]. 1. bracketleftBig , π bracketrightBig 2. bracketleftBig , 2 π bracketrightBig correct 3. bracketleftBig , 1 2 π bracketrightBig , bracketleftBig 3 2 π, 2 π bracketrightBig 4. bracketleftBig π, 2 π bracketrightBig 5. bracketleftBig 1 2 π, 3 2 π bracketrightBig Explanation: After differentiation, f ′ ( x ) = 1 − cos x. Now f will be increasing on an interval [ a, b ] if f ( x ) > 0 on ( a, b ). Determining where f ′ ( x ) > 0 can be done graphically or alge- braically. But the graph 1 2 π 2 π of 1 − cos x shows that f ′ ( x ) > 0 on the interval (0 , 2 π ). Thus f will be increasing everywhere on [0 , 2 π ] . On the other hand, we know that the inequal- ity cos x < 1 holds for all x on [0 , 2 π ] except when cos x = 1, i.e. , when x = 0 , 2 π. Thus 1 − cos x > 0 holds everywhere on (0 , 2 π ), showing algebraically that f will be increas- ing everywhere on [0 , 2 π ] . Version 020 – K Exam 2 – ben-zvi – (55600) 2 003 10.0 points Determine lim x →∞ 5 x 2 − 2 x + 7 5 + 5 x − 7 x 2 . 1. limit = ∞ 2. limit = 0 3. limit = 5 14 4. limit = − 5 7 correct 5. none of the other answers Explanation: Dividing the numerator and denominator by x 2 we see that 5 x 2 − 2 x + 7 5 + 5 x − 7 x 2 = 5 − 2 x + 7 x 2 5 x 2 + 5 x − 7 . On the other hand, lim x →∞ 1 x = lim x →∞ 1 x 2 = 0 . By Properties of limits, therefore, the limit = − 5 7 . 004 10.0 points Which of the following is the graph of f ( x ) = x − 2 x − 1 ? 1. 2 4 6 − 2 − 4 − 6 2 4 6 − 2 − 4 − 6 2. 2 4 6 − 2 − 4 − 6 2 4 6 − 2 − 4 − 6 3. 2 4 6 − 2 − 4 − 6 2 4 6 − 2 − 4 − 6 4. 2 4 6 − 2 − 4 − 6 2 4 6 − 2 − 4 − 6 5. 2 4 6 − 2 − 4 − 6 2 4 6 − 2 − 4 − 6 correct Version 020 – K Exam 2 – ben-zvi – (55600) 3 6. 2 4 6 − 2 − 4 − 6 2 4 6 − 2 − 4 − 6 Explanation: The graph of f will have y = 1 as a hori- zontal asymptote, and x − 1 = 0 as vertical asymptote. Combining these with the fact that the y-intercept of the graph occurs at y = +2, we see that the the graph of f must be 2 4 6 − 2 − 4 − 6 2 4 6 − 2 − 4 − 6 005 10.0 points The figure below shows the graphs of three functions:...
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K Exam 2-solutions-1 - Version 020 – K Exam 2 – ben-zvi...

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