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Pre-Calc Homework Solutions 68

# Pre-Calc Homework Solutions 68 - 68 Chapter 2 Review 48 At...

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42. One possible answer: 43. } f ( p p /2 /2 ) – f 0 (0) } 5 } 2 p 2 /2 1 } 5 } p 2 } 44. lim h 0 } V ( a 1 h h ) 2 V ( a ) } 5 lim h 0 5 } 1 3 } p H lim h 0 5 } 1 3 } p H lim h 0 (2 a 1 h ) 5 } 1 3 } p H (2 a ) 5 } 2 3 } p aH 45. lim h 0 5 lim h 0 5 lim h 0 5 lim h 0 (12 a 1 6 h ) 5 12 a 46. lim h 0 5 lim h 0 5 lim h 0 5 lim h 0 } 2 ah 1 h h 2 2 h } 5 lim h 0 (2 a 1 h 2 1) 5 2 a 2 1 47. (a) lim h 0 } f (1 1 h h ) 2 f (1) } 5 lim h 0 5 lim h 0 5 lim h 0 ( 2 1 1 h ) 5 2 1 ( b) The tangent at P has slope 2 1 and passes through (1, 2 2). y 5 2 1( x 2 1) 2 2 y 5 2 x 2 1 ( c) The normal at P has slope 1 and passes through (1, 2 2). y 5 1( x 2 1) 2 2 y 5 x 2 3 48. At x 5 a , the slope of the curve is lim h 0 } f ( a 1 h h ) 2 f ( a ) } 5 lim h 0 5 lim h 0 5 lim h 0 } 2 ah 2 h 3 h 1 h 2 } 5 lim h 0 (2 a 2 3 1 h ) 5 2 a 2 3 The tangent is horizontal when 2 a 2 3 5 0, at a 5 } 3 2 } 1 or x 5 } 3 2 } 2 . Since f 1 } 3 2 } 2 5 2 } 9 4 } , the point where this occurs is 1 } 3 2 } , 2 } 9 4 } 2 . 49. (a) p (0) 5 } 1 1 7 2 e 0 2 0 0.1(0) } 5 } 20 8 0 } 5 25 Perhaps this is the number of bears placed in the reserve when it was established.
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