Exam02-solutions

Exam02-solutions - Version 074 – Exam02 – Gilbert...

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Unformatted text preview: Version 074 – Exam02 – Gilbert – (56215) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the interval(s) where f ( x ) = x 3 + 9 x 2 + 6 is increasing. 1. ( −∞ , − 3) , (3 , ∞ ) 2. ( −∞ , − 6] , [0 , ∞ ) correct 3. [ − 3 , 0] 4. ( −∞ , − 3] , [0 , ∞ ) 5. [ − 6 , 0] 6. ( −∞ , − 6) , (6 , ∞ ) Explanation: The function f will be increasing on an interval [ a, b ] (resp. ( −∞ , b ] or [ a, ∞ )) when f ′ > 0 on ( a, b ) (resp. ( −∞ , b ) or ( a, ∞ )), i.e. , on those values of x for which f ′ ( x ) = 3 x 2 + 18 x = 3 x ( x + 6) > . Consequently, f will be increasing on ( −∞ , − 6] , [0 , ∞ ) . 002 10.0 points Find all asymptotes of the graph of y = 3 x 2 − 16 x + 16 x 2 − 5 x + 4 . 1. vertical: x = 1 , horizontal: y = − 3 2. vertical: x = 4 , horizontal: y = − 3 3. vertical: x = 4 , horizontal: y = 3 4. vertical: x = 1 , 4 , horizontal: y = 3 5. vertical: x = 1 , horizontal: y = 3 cor- rect 6. vertical: x = 1 , 4 , horizontal: y = − 3 7. vertical: x = − 1 , horizontal: y = 3 Explanation: After factorization y = (3 x − 4)( x − 4) ( x − 1)( x − 4) . Thus y is not defined at x = 4, but for x negationslash = 4 y = 3 x − 4 x − 1 ; notice, however, that lim x → 4 3 x − 4 x − 1 = 8 3 exists, so the graph does not have a vertical asymptote at x = 4. Since vextendsingle vextendsingle vextendsingle 3 x − 4 x − 1 vextendsingle vextendsingle vextendsingle −→ ∞ as x → 1 from the left and the right, the line x = 1 will, however, be a vertical asymptote. On the other hand, lim x →±∞ 3 x − 4 x − 1 = 3 , so y = 3 will be a horizontal aymptote. Hence the graph has only the following asymptotes vertical: x = 1 , horizontal: y = 3. 003 10.0 points Determine which of the following graphs could be that of f ( x ) = 3 + 2 x 2 1 + x 2 . Version 074 – Exam02 – Gilbert – (56215) 2 1....
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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.

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Exam02-solutions - Version 074 – Exam02 – Gilbert...

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