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Pre-Calc Homework Solutions 284

# Pre-Calc Homework Solutions 284 - e 2 x 2 Since e 2 x 2 0...

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27. } d d y x } 5 y 1 2 } y d 1 y 2 } 5 dx E } y d 1 y 2 } 5 E dx ln ) y 1 2 ) 5 x 1 C y 1 2 5 Ce x y 5 Ce x 2 2 y (0) 5 C 2 2 5 2 C 5 4 y 5 4 e x 2 2 Graphical support: [ 2 5, 5] by [ 2 5, 20] 28. } d d y x } 5 (2 x 1 1)( y 1 1) } y d 1 y 1 } 5 (2 x 1 1) dx E } y d 1 y 1 } 5 E (2 x 1 1) dx ln ) y 1 1 ) 5 x 2 1 x 1 C y 1 1 5 Ce x 2 1 x y 5 Ce x 2 1 x 2 1 y ( 2 1) 5 C 2 1 5 1 C 5 2 y 5 2 e x 2 1 x 2 1 Graphical support: [ 2 3, 3] by [ 2 10, 40] 29. E 2 f ( x ) dx 5 2 E f ( x ) dx 5 2 (1 2 ˇ x w ) 1 C 5 2 1 1 ˇ x w 1 C Since 2 1 1 C is an arbitrary constant, we may write the indefinite integral as ˇ x w 1 C . 30. E [ x 1 f ( x )] dx 5 E x dx 1 E f ( x ) dx 5 } x 2 2 } 1 (1 2 ˇ x w ) 1 C 5 } x 2 2 } 1 1 2 ˇ x w 1 C Since 1 1 C is an arbitrary constant, we may write the indefinite integral as } x 2 2 } 2 ˇ x w 1 C . 31. E [2 f ( x ) 2 g ( x )] dx 5 2 E f ( x ) dx 2 E g ( x ) dx 5 2(1 2 ˇ x w ) 2 ( x 1 2) 1 C 5 2 2 ˇ x w 2 x 1 C 32. E [ g ( x ) 2 4] dx 5 E g ( x ) dx 2 E 4 dx 5 ( x 1 2) 2 4 x 1 C 5 2 2 3 x 1 C Since 2 1 C is an arbitrary constant, we may write the indefinite integral as 2 3 x 1 C . 33. We seek the graph of a function whose derivative is } sin x x } . Graph (b) is increasing on [ 2 p , p ], where } sin x x } is positive, and oscillates slightly outside of this interval. This is the correct choice, and this can be verified by graphing NINT 1 } sin x x } , x , 0, x 2 . 34. We seek the graph of a function whose derivative is
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Unformatted text preview: e 2 x 2 . Since e 2 x 2 . 0 for all x , the desired graph is increasing for all x . Thus, the only possibility is graph (d), and we may verify that this is correct by graphing NINT( e 2 x 2 , x , 0, x ). 35. (iv) The given graph looks like the graph of y 5 x 2 , which satisfies } d d y x } 5 2 x and y (1) 5 1. 36. Yes, y 5 x is a solution. 37. (a) } d d v t } 5 2 1 6 t E dv 5 E (2 1 6 t ) dt v 5 2 t 1 3 t 2 1 C Initial condition: v 5 4 when t 5 4 5 1 C 4 5 C v 5 2 t 1 3 t 2 1 4 (b) E 1 v ( t ) dt 5 E 1 (2 t 1 3 t 2 1 4) dt 5 3 t 2 1 t 3 1 4 t 4 5 6 2 5 6 The particle moves 6 m. 1 284 Chapter 6 Review...
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