thomasET_226348_ism08

thomasET_226348_ism08 - 32 Chapter 1 Functions 1.5...

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32 Chapter 1 Functions 1.5 EXPONENTIAL FUNCTIONS 1. 2. 3. 4. 5. 6. 7. 8.
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Section 1.5 Exponential Functions 33 9. 10. 11. 16 16 16 16 16 2 2 1.75 2 1.75 0.25 1 4 †œ œ œ œ ±² ± Î ab 12. 9 9 9 9 3 13 16 12 ÎÎ Î œœ 11 36 13. 4 4 4 2 4 4 4.2 3.7 0.5 1 2 4.2 3.7 œ œ ±Î 14. 3 3 3 3 3 1 53 23 52 33 Î Î œ 15. 25 25 25 5 ˆ‰ 18 48 4 Î œœœ 16. 13 13 13 Š‹ È È 22 2 Î Î 17. 2 7 2 7 14 ÈÈ È È 3 3 œ 18. 3 12 3 12 36 6 Š Š È Î Î Î œ† œ œ 19. 4 1 6 2 4 2 2 È œ 4 42 Î 20. Ȉ 6 9 3 2 6 62 œ 2 2 Î 21. Domain: , ; y in range y . As x increases, e becomes infinitely large and y becomes a smaller and ±_ _ Ê œ 1 2e x ² x smaller positive real number. As x decreases, e becomes a smaller and smaller positive real number, y , and y gets x ² " # arbitrarily close to Range: , . "" ## Ê! 22. Domain: , ; y in range y cos e . Since the values of e are , and 1 cos x 1 Range: 1, 1 . a b a b ±_ _ Ê œ ! _ ± Ÿ Ÿ Ê Ò± Ó ±± tt 23. Domain: , ; y in range y 1 3 . Since the values of 3 are , Range: 1, . a b È ±_ _ Ê œ ³ ! _ Ê _ ± ± t t 24. If e 1, then x Domain: , , ; y in range y . If x , then 1 e 2x 2x 3 1e ! Ê ± _ ! ´ ! _ Ê œ µ ! ² ² _ a b ± 2x y . If x , then e 1 3 y Range: , 3, . ʱ_² ²! ²! ² Ê ² ²_Ê ±_ ! ´ _ 2x a b
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34 Chapter 1 Functions 25. 26. x 2.3219 x 1.3863 ¸¸ 27. 28. x 0.6309 x 1.5850 ¸± 29. Let t be the number of years. Solving 500,000 1.0375 1,000,000 graphically, we find that t 18.828. The population ab t œ¸ will reach 1 million in about 19 years. 30. (a) The population is given by P t 6250 1.0275 , where t is the number of years after 1890. a b œ t Population in 1915: P 25 12,315 ¸ Population in 1940: P 50 24,265 ¸ (b) Solving P t 50,000 graphically, we find that t 76.651. The population reached 50,000 about 77 years after 1890, in 1967. 31. (a) A t 6.6 ˆ‰ œ 1 2 t14 Î (b) Solving A t 1 graphically, we find that t 38. There will be 1 gram remaining after about 38.1145 days. 32. Let t be the number of years. Solving 2300 1.06 4150 graphically, we find that t 10.129. It will take about 10.129 t years. (If the interest is not credited to the account until the end of each year, it will take 11 years.) 33. Let A be the amount of the initial investment, and let t be the number of years. We wish to solve A 1.0625 2A, which t œ is equivalent to 1.0625 2. Solving graphically, we find that t 11.433. It will take about 11.433 years. (If the interest is t credited at the end of each year, it will take 12 years.) 34. Let A be the amount of the initial investment, and let t be the number of years. We wish to solve A 1 2A, ²œ 0.0625 12 12t which is equivalent to 1 2. Solving graphically, we find that t 11.119. It will take about 11.119 years. (If ¸ 0.0625 12 12t the interest is credited at the end of each month, it will take 11 years 2 months.) 35. Let A be the amount of the initial investment, and let t be the number of years. We wish to solve Ae 2A, which is 0.0625t œ equivalent to e 2. Solving graphically, we find that t 11.090. It will take about 11.090 years. 0.0625t
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Section 1.6 Inverse Functions and Logarithms 35 36. Let A be the amount of the initial investment, and let t be the number of years. We wish to solve A 1.0575 3A, which ab t œ is equivalent to 1.0575 3. Solving graphically, we find that t 19.650. It will take abou 19.650 years. (If the interest t œ¸ is credited at the end of each year, it will take 20 years.) 37. Let A be the amount of the initial investment, and let t be the number of years. We wish to solve A 1
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thomasET_226348_ism08 - 32 Chapter 1 Functions 1.5...

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