Business Calc Homework w answers_Part_3

Business Calc Homework w answers_Part_3 - Section 1.2 55....

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55. [ 2 5, 5] by [ 2 2, 5] We require x 2 2 4 $ 0 (so that the square root is defined) and x 2 2 4 ± 0 (to avoid division by zero), so the domain is ( 2‘ , 2 2) < (2, ). For values of x in the domain, x 2 2 4 1 and hence ˇ x 2 w 2 w 4 w and } ˇ x 2 w 1 2 w 4 w } 2 can attain any positive value, so the range is (0, ). (Note that grapher failure may cause the range to appear as a finite interval on a grapher. 56. [ 2 5, 5] by [ 2 2, 5] We require 9 2 x 2 $ 0 (so that the fourth root is defined) and 9 2 x 2 ± 0 (to avoid division by zero), so the domain is ( 2 3, 3). For values of x in the domain, 9 2 x 2 can attain any value in (0, 9]. Therefore, ˇ 4 9 w 2 w x w 2 w can attain any value in (0, ˇ 3 w ], and } ˇ 4 9 w 2 2 w x w 2 w } can attain any value in 3 } ˇ 2 3 w } , 2 . The range is 3 } ˇ 2 3 w } , 2 or approximately [1.15, ). (Note that grapher failure may cause the range to appear as a finite interval on a grapher.) 57. [ 2 4.7, 4.7] by [ 2 3.1, 3.1] We require 9 2 x 2 ± 0, so the domain is ( 2‘ , 2 3) < ( 2 3, 3) < (3, ). For values of x in the domain, 9 2 x 2 can attain any value in ( 2‘ , 0) < (0, 9], so ˇ 3 9 w 2 w x w 2 w can attain any value in ( 2‘ , 0) < (0, ˇ 3 9 w ]. Therefore, } ˇ 3 9 w 2 2 w x w 2 w } can attain any value in ( 2‘ , 0) < 3 } ˇ 3 2 9 w } , 2 . The range is ( 2‘ , 0) < 3 } ˇ 3 2 9 w } , 2 or approximately ( 2‘ , 0) < [0.96, ). (Note that grapher failure can cause the intervals in the range to appear as finite intervals on a grapher.) 58. [ 2 2.35, 2.35] by [ 2 1.55, 1.55] We require x 2 2 1 ± 0, so the domain is ( 2‘ , 2 1) < ( 2 1, 1) < (1, ). For values of x in the domain, x 2 2 1 can attain any value in [ 2 1, 0) < (0, ), so ˇ 3 x 2 w 2 w 1 w can also attain any value in [ 2 1, 0) < (0, ). Therefore, } ˇ 3 x 1 w 2 w 1 w } can attain any value in ( 2‘ , 2 1] < (0, ). The range is ( 2‘ , 2 1] < (0, ). (Note that grapher failure can cause the intervals in the range to appear as finite intervals on a grapher.) 59. (a) (b) 60. (a) y x 2 2 1 –1 –1 –2 –2 y x 1.5 2 0 –2 –1.5 y x 1.5 2 0 –2 –1.5 Section 1.2 11
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60. continued (b) 61. (a) (b) 62. (a) (b) 63. (a) Since ( f + g )( x ) 5 ˇ g w ( x w ) w 2 w 5 w 5 ˇ x 2 w 2 w 5 w , g ( x ) 5 x 2 . (b) Since ( f + g )( x ) 5 1 1 } g ( 1 x ) } 5 x , we know that } g ( 1 x ) } 5 x 2 1, so g ( x ) 5 } x 2 1 1 } . (c) Since ( f + g )( x ) 5 f 1 } 1 x } 2 5 x , f ( x ) 5 } 1 x } . (d) Since ( f + g )( x ) 5 f ( ˇ x w ) 5 ) x ) , f ( x ) 5 x 2 . The completed table is shown. Note that the absolute value sign in part (d) is optional. 64. (a) Note that the data in the table begins at x 5 20. (We do not include the initial investment in the data.) The power regression equation is y 5 27.1094 x 2.651044 .
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This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.

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Business Calc Homework w answers_Part_3 - Section 1.2 55....

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