Business Calc Homework w answers_Part_7

Business Calc Homework w answers_Part_7 - 31 Section 1.6...

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50. (a) y 52 590.969 1 152.817 ln x (b) When x 5 85, y < 87.94. About 87.94 million metric tons were produced. (c) 2 590.969 1 152.817 ln x 5 120 152.817 ln x 5 710.969 ln x 5 } 7 1 1 5 0 2 . . 9 8 6 1 9 7 } x 5 e } 7 1 1 5 0 2 . . 9 8 6 1 9 7 } < 104.84 According to the regression equation, oil production will reach 120 million metric tons when x < 104.84, in about 2005. 51. (a) Suppose that f ( x 1 ) 5 f ( x 2 ). Then mx 1 1 b 5 mx 2 1 b so mx 1 5 mx 2 . Since m ± 0, this gives x 1 5 x 2 . (b) y 5 mx 1 b y 2 b 5 mx } y 2 m b } 5 x Interchange x and y . } x 2 m b } 5 y f 2 1 ( x ) 5 } x 2 m b } The slopes are reciprocals. (c) If the original functions both have slope m , each of the inverse functions will have slope } m 1 } . The graphs of the inverses will be parallel lines with nonzero slope. (d) If the original functions have slopes m and 2} m 1 } , respectively, then the inverse functions will have slopes } m 1 } and 2 m , respectively. Since each of } m 1 } and 2 m is the negative reciprocal of the other, the graphs of the inverses will be perpendicular lines with nonzero slopes. 52. (a) y 2 is a vertical shift (upward) of y 1 , although it’s difficult to see that near the vertical asymptote at x 5 0. One might use “trace” or “table” to verify this. (b) Each graph of y 3 is a horizontal line. (c) The graphs of y 4 and y 5 a are the same. (d) e y 2 2 y 1 5 a , ln( e y 2 2 y 1 ) 5 ln a , y 2 2 y 1 5 ln a , y 1 5 y 2 2 ln a 5 ln x 2 ln a 53. If the graph of f ( x ) passes the horizontal line test, so will the graph of g ( x ) f ( x ) since it’s the same graph reflected about the x -axis. Alternate answer: If g ( x 1 ) 5 g ( x 2 ) then 2 f ( x 1 ) f ( x 2 ), f ( x 1 ) 5 f ( x 2 ), and x 1 5 x 2 since f is one-to-one. 54. Suppose that g ( x 1 ) 5 g ( x 2 ). Then } f ( 1 x 1 ) } 5 } f ( 1 x 2 ) } , f ( x 1 ) 5 f ( x 2 ), and x 1 and x 2 since f is one-to-one. 55. (a) The expression a ( b c 2 x ) 1 d is defined for all values of x , so the domain is ( 2‘ , ). Since b c 2 x attains all positive values, the range is ( d , ) if a . 0 and the range is ( 2‘ , d ) if a , 0. (b) The expression a log b ( x 2 c ) 1 d is defined when x 2 c . 0, so the domain is ( c , ). Since a log b ( x 2 c ) 1 d attains every real value for some value of x , the range is ( 2‘ , ). 56. (a) Suppose f ( x 1 ) 5 f ( x 2 ). Then: } a cx x 1 1 1 1 d b } 5 } a cx x 2 2 1 1 d b } ( ax 1 1 b )( cx 2 1 d ) 5 ( ax 2 1 b )( cx 1 1 d ) acx 1 x 2 1 adx 1 1 bcx 2 1 bd 5 acx 1 x 2 1 adx 2 1 bcx 1 1 bd adx 1 1 bcx 2 5 adx 2 1 bcx 1 ( ad 2 bc ) x 1 5 ( ad 2 bc ) x 2 Since ad 2 bc ± 0, this means that x 1 5 x 2 . (b) y 5 } a cx x 1 1 d b } cxy 1 dy 5 ax 1 b ( cy 2 a ) x dy 1 b x 5 } 2 c d y y 2 1 a b } Interchange x and y: y 5 } 2 c d x x 2 1 a b } f 2 1 ( x ) 5 } 2 c d x x 2 1 a b } (c) As x 6‘ , f ( x ) 5 } a cx x 1 1 d b } } a c } , so the horizontal asymptote is y 5 } a c } ( c ± 0). Since f ( x ) is undefined at x 52} d c } , the vertical asymptote is x d c } . (d) As x 6‘ , f 2 1 ( x ) 5 } 2 c d x x 2 1 a b } 2} d c } , so the horizon- tal asymptote is y d c } ( c ± 0). Since f 2 1 ( x ) is undefined at x 5 } a c } , the vertical asymptote is x 5 } a c } .
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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_7 - 31 Section 1.6...

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