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Unformatted text preview: husain (aih243) – Exam02Review – Gilbert – (56215) 1 This printout should have 16 questions. Multiplechoice 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 − + −∞ ∞ 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 xaxis. Hence the graph of f must lie below the xaxis 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: 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....
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This note was uploaded on 11/30/2010 for the course M 408c taught by Professor Mcadam during the Spring '06 term at University of Texas.
 Spring '06
 McAdam

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