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ReviewExam02-sol - husain(aih243 Exam02Review Gilbert(56215...

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husain (aih243) – Exam02Review – Gilbert – (56215) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points On which interval(s) is f ( x ) = x 4 + 2 x 2 1 increasing? 1. ( −∞ , 1 ] , [ 1 , ) 2. ( −∞ , 0 ] 3. [ 1 , 1 ] 4. [ 0 , ) correct 5. [ 1 , 0 ] , [ 1 , ) 6. ( −∞ , 1 ] , [ 0 , 1 ] Explanation: A function f is increasing on an interval [ a , b ] when f ( x ) > 0 on ( a, b ). Now f ( x ) = 4 x 3 + 4 x = 4 x ( x 2 + 1) . From the sign chart + 0 −∞ for f it thus follows that f is increasing on [ 0 , ) . 002 10.0 points A polynomial function f has the properties (i) f ( 8) = 0 = f ( 2), (ii) f ′′ ( x ) > 0 on ( 8 , 2). Which of the following statements is a conse- quence of these properties? 1. f ( x ) 0 on [ 8 , 2] 2. f ( x ) 0 on ( 8 , 2) 3. f ( x ) = 0 on [ 8 , 2] 4. f ( x ) 0 on [ 8 , 2] correct 5. f ( x ) 0 on ( 8 , 2) Explanation: Condition (ii) ensures that the graph of f is concave up on the interval ( 8 , 2). On the other hand, by (i), f ( 8) = 0 = f ( 2), so both endpoints of the graph on [ 8 , 2] lie on the x -axis. Hence the graph of f must lie below the x -axis on the interval ( 8 , 2), i.e. , f ( x ) 0 on [ 8 , 2] . Notice that the inequality f ( x ) < 0 must hold on the interval ( 8 , 2) because f ′′ ( x ) > 0 on ( 8 , 2). 003 10.0 points Which one of the following properties does f ( x ) = x 1 x 2 + 24 have? 1. local max at x = 6 2. local max at x = 4 3. local min at x = 6 4. local min at x = 4 5. local min at x = 6 6. local max at x = 6 correct Explanation:
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husain (aih243) – Exam02Review – Gilbert – (56215) 2 By the Quotient rule, f ( x ) = x 2 + 24 2 x ( x 1) ( x 2 + 24) 2 = 24 + 2 x x 2 ( x 2 + 24) 2 . The critical points of f occur when f ( x ) = 0, i.e. , at the solutions of f ( x ) = (4 + x )(6 x ) ( x 2 + 24) 2 = 0 . Thus the critical points of f are x = 4 and x = 6. To classify these critical points we use the First Derivative Test. But the sign of f de- pends only on the numerator, so it is enough, therefore, to look only at a sign chart for (4 + x )(6 x ): 4 6 + From this it follows that f is decreasing on ( −∞ , 4), increasing on ( 4 , 6), and de- creasing on (6 , ). Consequently, f has a local maximum at x = 6 . keywords: local maximum, local minimum, critical point, quotient rule, First Derivative Test, rational function 004 10.0 points Determine if the limit lim x → −∞ 2 + x 5 x 3 2 + 4 x 3 exists, and if it does, find its value. 1. limit = 1 2. limit does not exist 3. limit = 1 4. limit = 5 4 5. limit = 5 4 correct Explanation: Dividing by x 3 in the numerator and de- nominator we see that 2 + x 5 x 3 2 + 4 x 3 = 2 x 3 + 1 x 2 5 2 x 3 + 4 .
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