Business Calc Homework w answers_Part_35

Business Calc Homework w answers_Part_35 - Section 4.5 13...

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13. Center 5 1 f 9 ( x ) 5 } ( x 1 1 ( x )( 1 1) 1 2 ) 2 ( x )(1) }5 } ( x 1 1 1) 2 } We have f (1) 5 } 1 2 } and f 9 (1) 5 } 1 4 } L ( x ) 5 f (1) 1 f 9 (1)( x 2 1) 5 } 1 2 } 1 } 1 4 } ( x 2 1) 5 } 1 4 } x 1 } 1 4 } Alternate solution: Using center 5 } 3 2 } , we have f 1 } 3 2 } 2 5 } 3 5 } and f 9 1 } 3 2 } 2 5 } 2 4 5 } . L ( x ) 5 f 1 } 3 2 } 2 1 f 9 1 } 3 2 } 21 x 2 } 3 2 } 2 5 } 3 5 } 1 } 2 4 5 } 1 x 2 } 3 2 } 2 5 } 2 4 5 } x 1} 2 9 5 } 14. Center 5 } p 2 } f 9 ( x ) 52 sin x We have f 1 } p 2 } 2 5 0 and f 9 1 } p 2 } 2 1. L ( x ) 5 f 1 } p 2 } 2 1 f 9 1 } p 2 } 21 x 2 } p 2 } 2 5 0 2 1 1 x 2 } p 2 } 2 x 1 } p 2 } 15. Let f ( x ) 5 x 3 1 x 2 1. Then f 9 ( x ) 5 3 x 2 1 1 and x n 1 1 5 x n 2 } f f 9 ( ( x x n n ) ) } 5 x n 2 } x n 3 3 x 1 n 2 x 1 n 2 1 1 } . Note that f is cubic and f 9 is always positive, so there is exactly one solution. We choose x 1 5 0. x 1 5 0 x 2 5 1 x 3 5 0.75 x 4 < 0.6860465 x 5 < 0.6823396 x 6 < 0.6823278 x 7 < 0.6823278 Solution: x < 0.682328 16. Let f ( x ) 5 x 4 1 x 2 3. Then f 9 ( x ) 5 4 x 3 1 1 and x n 1 1 5 x n 2 } f f 9 ( ( x x n n ) ) } 5 x n 2 } x n 4 4 x 1 n 3 x 1 n 2 1 3 } The graph of y 5 f ( x ) shows that f ( x ) 5 0 has two solutions. [ 2 3, 3] by [ 2 4, 4] x 1 1.5 x 1 5 1.2 x 2 1.455 x 2 < 1.6541962 x 3 < 2 1.4526332 x 3 < 1.1640373 x 4 < 2 1.4526269 x 4 < 1.1640351 x 5 < 2 1.4526269 x 5 < 1.1640351 Solutions: x < 2 1.452627, 1.164035 17. Let f ( x ) 5 x 2 2 2 x 1 1 2 sin x . Then f 9 ( x ) 5 2 x 2 2 2 cos x and x n 1 1 5 x n 2 } f f 9 ( ( x x n n ) ) } 5 x n 2 The graph of y 5 f ( x ) shows that f ( x ) 5 0 has two solutions [ 2 4, 4] by [ 2 3, 3] x 1 5 0.3 x 1 5 2 x 2 < 0.3825699 x 2 < 1.9624598 x 3 < 0.3862295 x 3 < 1.9615695 x 4 < 0.3862369 x 4 < 1.9615690 x 5 < 0.3862369 x 5 < 1.9615690 Solutions: x < 0.386237, 1.961569 18. Let f ( x ) 5 x 4 2 2. Then f 9 ( x ) 5 4 x 3 and x n 1 1 5 x n 2 } f f 9 ( ( x x n n ) ) } 5 x n 2 } x n 4 4 x 2 n 3 2 } . Note that f ( x ) 5 0 clearly has two solutions, namely x 56 ˇ 4 2 w . We use Newton’s method to find the decimal equivalents. x 1 5 1.5 x 2 < 1.2731481 x 3 < 1.1971498 x 4 < 1.1892858 x 5 < 1.1892071 x 6 < 1.1892071 Solutions: x < 6 1.189207 19. (a) Since } d d y x } 5 3 x 2 2 3, dy 5 (3 x 2 2 3) dx . (b) At the given values, dy 5 (3 ? 2 2 2 3)(0.05) 5 9(0.05) 5 0.45. 20. (a) Since } d d y x } 55 } ( 2 1 2 1 2 x x 2 ) 2 2 } , dy 5 } ( 2 1 2 1 2 x x 2 ) 2 2 } dx . (b) At the given values, dy 5 } [ 2 1 2 1 2 ( 2 ( 2 2 2 ) 2 ) ] 2 2 } (0.1) 5 } 2 5 2 2 8 } (0.1) 0.024. 21. (a) Since } d d y x } 5 ( x 2 ) 1 } 1 x } 2 1 (ln x )(2 x ) 5 2 x ln x 1 x , dy 5 (2 x ln x 1 x ) dx . (b) At the given values, dy 5 [2(1) ln (1) 1 1](0.01) 5 1(0.01) 5 0.01 (1 1 x 2 )(2) 2 (2 x )(2 x ) }}} (1 1 x 2 ) 2 x n 2 2 2 x n 1 1 2 sin x n }}} 2 x n 2 2 2 cos x n Section 4.5 171
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22. (a) Since } d d y x } 5 ( x ) 1 } 2 ˇ 1 w 1 2 w x w 2 w } 2 ( 2 2 x ) 1 ( ˇ 1 w 2 w x w 2 w )(1) 5 } ˇ 1 w 2 2 w x 2 x w 2 w }1 ˇ 1 w 2 w x w 2 w 5 } 2 x 2 ˇ 1 1 w ( 2 w 1 2 x w 2 w x 2 ) }5 } ˇ 1 1 w 2 2 w 2 x x w 2 2 w } , dy 5 } ˇ 1 1 w 2 2 w 2 x x w 2 2 w } dx . (b) At the given values, dy 5 } ˇ 1 1 w 2 2 w 2( ( w 0 0 w ) ) 2 2 w } ( 2 0.2) 52 0.2. 23. (a) Since } d d y x } 5 e sin x cos x , dy 5 (cos x ) e sin x dx . (b) At the given values, dy 5 (cos p )( e sin p )( 2 0.1) 5 ( 2 1)(1)( 2 0.1) 5 0.1. 24. (a) Since } d d y x }52 3 csc 1 1 2 } 3 x } 2 cot 1 1 2 } 3 x } 21 2} 1 3 } 2 5 csc 1 1 2 } 3 x } 2 cot 1 1 2 } 3 x } 2 , dy 5 csc 1 1 2 } 3 x } 2 cot 1 1 2 } 3 x } 2 dx . (b) At the given values, dy 5 csc 1 1 2 } 1 3 } 2 cot 1 1 2 } 1 3 } 2 (0.1) 5 0.1 csc } 2 3 } cot } 2 3 } < 0.205525 25. (a) y 1 xy 2 x 5 0 y (1 1 x ) 5 x y 5 } x 1 x 1 } Since } d d y x } 5 } ( x 1 1 ( x )( 1 1) 1 2 ) 2 ( x )(1) } ( x 1 1 1) 2 } , dy 5 } ( x 1 dx 1) 2 } .
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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_35 - Section 4.5 13...

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