# Exam1 Review - f e Â 2 1[0 1 domain range of f Â 1[0 1 e Â 2 1 11(a 3 p 7 7(b x p 4 Â x 2 2 12 a 1 0(2 1 b(0;e 3 = 2 e 3 = 2 1 13 â€¦ 6 5 â€¦ 6 3

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Review Answers, Exam 1, Spring 2012 Part I: Multiple Choice Answers 1. d 2. e 3. a 4. a 5. e 6. d 7. a 8. d 9. c 10. a 11. c 12. e 13. a 14. b 15. d 16. a 17. none: ( ¡1 ; ¡ 2] [ (0 ; 2] 18. b and d 19. c 20. b Part II: Free Response Answers 1. a. ( ¡1 ; ¡ 2), ( ¡ 2 ; 0], (0 ; 3), (3 ; 4), (4 ; 1 ) b. 0 c. 4 d. ¡ 2, 3 e. f ( ¡ 2) = 0 and f (3) = 0

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2. a. 5 4 b. ¡ 6 c. 0 d. 1 3. a) 2 b) 1 2 c) + 1 d) ¡ 1 x = ¡ 1: removable, x = 0: removable, x = 1: inﬂnite vertical asymptote: x = 1, horizontal asymptote: y = ¡ 1 4. a) -4 y 3 -1 1 2 2 0 3 4 -2 1 -3 0 -4 -2 -1 x -3 4 b) graph of y = p x is shifted right three units, re±ected over the x -axis, and then shifted up 2 units 5. x = 3 ; 5 3 6. ¡ 2 p 5 ¡ 2 3 p 5 7. a. M ( t ) = 50 ± 1 2 t= 60 b. t = 60ln ± 3 10 ln ± 1 2 104.2 days 8. (a) ¡ 1 (b) ¡ 4 (c) 0; use Squeeze Theorem (d) 1) 0 2) ¡ 2 3 3) horizontal asymptotes: y = ¡ 2 3 and y = 0, vertical asymptote: x = ¡ ln3 9. a) does not exist b) ¡1 f ( x ) has a jump discontinuity at x = 0 and an inﬂnite discontinuity at x = 2.
10. f ¡ 1 ( x ) = e x 2 ¡ 2 domain, range of
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Unformatted text preview: f : [ e Â¡ 2 ; 1 ), [0 ; 1 ) domain, range of f Â¡ 1 : [0 ; 1 ), [ e Â¡ 2 ; 1 ) 11. (a) 3 p 7 7 (b) x p 4 Â¡ x 2 2 12. a. ( 1 ; 0) [ (2 ; 1 ) b. (0 ;e 3 = 2 ) [ ( e 3 = 2 ; 1 ) 13. â€¦ 6 , 5 â€¦ 6 , 3 â€¦ 2 14. (a) Â¡ 2 ( x + h Â¡ 2)( x Â¡ 2) (b) Â¡ 2 ( x Â¡ 2) 2 15. x = 1 only 16. (b) 2, 2, 2, 1, 0, Â¡1 (c) x = 0: removable, deï¬‚ne f (0) = 2, x = 2 Â¡ : removable, deï¬‚ne f (2) = 0 to make f continuous from the left at x = 2, x = 2 + : inï¬‚nite, so x = 2: nonremovable 17. A ( x ) = x Â± 1500 Â¡ 3 2 x Â¶ If area is 375,000 sq. ft., the dimensions are 500 ft by 750 ft. with x = 500. 18. Area: A ( x ) = x 2 ( Â¡ â€¦ 8 Â¡ 1 2 ) + 10 x 19. (a) 16 ft/sec (b) Â¡ 32 Â¡ 16 h ft/sec (c) Â¡ 32 ft/sec...
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## This note was uploaded on 02/14/2012 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

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Exam1 Review - f e Â 2 1[0 1 domain range of f Â 1[0 1 e Â 2 1 11(a 3 p 7 7(b x p 4 Â x 2 2 12 a 1 0(2 1 b(0;e 3 = 2 e 3 = 2 1 13 â€¦ 6 5 â€¦ 6 3

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