Calculus: Early Transcendentals, by Anton, 7th Edition,ch04

Calculus - Early Transcendentals

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122 CHAPTER 4 Exponential, Logarithmic, and Inverse Trigonometric Functions EXERCISE SET 4.1 1. (a) f ( g ( x )) = 4( x/ 4) = x , g ( f ( x )) = (4 x ) / 4= x , f and g are inverse functions (b) f ( g ( x )) = 3(3 x 1)+1=9 x 2 6 = x so f and g are not inverse functions (c) f ( g ( x )) = 3 p ( x 3 +2) 2= x , g ( f ( x )) = ( x 2)+2= x , f and g are inverse functions (d) f ( g ( x )) = ( x 1 / 4 ) 4 = x , g ( f ( x )) = ( x 4 ) 1 / 4 = | x | 6 = x , f and g are not inverse functions 2. (a) They are inverse functions. 2 -2 -2 2 (b) The graphs are not reflections of each other about the line y = x . 2 -2 -2 2 (c) They are inverse functions provided the domain of g is restricted to [0 , + ) 5 0 05 (d) They are inverse functions provided the domain of f ( x ) is restricted to [0 , + ) 2 0 02 3. (a) yes; all outputs (the elements of row two) are distinct (b) no; f (1) = f (6)
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Exercise Set 4.1 123 4. (a) no; it is easy to conceive of, say, 8 people in line at two diFerent times (b) no; perhaps your weight remains constant for more than a year (c) yes, since the function is increasing, in the sense that the greater the volume, the greater the weight 5. (a) yes (b) yes (c) no (d) yes (e) no (f) no 6. (a) no, the horizontal line test fails 6 -2 -3 3 (b) yes, horizontal line test 10 -10 -1 3 7. (a) no, the horizontal line test fails (b) no, the horizontal line test fails (c) yes, horizontal line test 8. (d) no, the horizontal line test fails (e) no, the horizontal line test fails yes, horizontal line test 9. (a) f has an inverse because the graph passes the horizontal line test. To compute f 1 (2) start at2onthe y -axis and go to the curve and then down, so f 1 (2) = 8; similarly, f 1 ( 1) = 1 and f 1 (0) = 0. (b) domain of f 1 is [ 2 , 2], range is [ 8 , 8] (c) -2 1 2 -8 -4 4 8 y x 10. (a) the horizontal line test fails (b) −∞ <x ≤− 1; 1 x 2; and 2 x< 4. 11. (a) f 0 ( x )=2 x +8; f 0 < 0on( −∞ , 4) and f 0 > 4 , + ); not one-to-one (b) f 0 ( x )=10 x 4 +3 x 2 3 > 0; f 0 ( x ) is positive for all x ,so f is one-to-one (c) f 0 ( x ) = 2 + cos x 1 > 0 for all x f is one-to-one 12. (a) f 0 ( x )=3 x 2 +6 x = x (3 x + 6) changes sign at x = 2 , 0, so f is not one-to-one (b) f 0 ( x )=5 x 4 +24 x 2 +2 2 > 0; f 0 is positive for all x f is one-to-one (c) f 0 ( x )= 1 ( x +1) 2 ; f is one-to-one because: if x 1 2 < 1 then f 0 > 0on[ x 1 ,x 2 ], so f ( x 1 ) 6 = f ( x 2 ) if 1 1 2 then f 0 > x 1 2 ], so f ( x 1 ) 6 = f ( x 2 ) if x 1 < 1 2 then f ( x 1 ) > 1 >f ( x 2 ) since f ( x ) > 1on( −∞ , 1) and f ( x ) < 1on ( 1 , + )
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124 Chapter 4 13. y = f 1 ( x ), x = f ( y )= y 5 , y = x 1 / 5 = f 1 ( x ) 14. y = f 1 ( x ), x = f ( y )=6 y , y = 1 6 x = f 1 ( x ) 15. y = f 1 ( x ), x = f ( y )=7 y 6, y = 1 7 ( x +6)= f 1 ( x ) 16. y = f 1 ( x ), x = f ( y y +1 y 1 , xy x = y +1,( x 1) y = x +1, y = x x 1 = f 1 ( x ) 17. y = f 1 ( x ), x = f ( y )=3 y 3 5, y = 3 p ( x +5) / 3= f 1 ( x ) 18. y = f 1 ( x ), x = f ( y 5 4 y +2, y = 1 4 ( x 5 2) = f 1 ( x ) 19. y = f 1 ( x ), x = f ( y 3 2 y 1, y =( x 3 +1) / 2= f 1 ( x ) 20. y = f 1 ( x ), x = f ( y 5 y 2 , y = r 5 x x = f 1 ( x ) 21. y = f 1 ( x ), x = f ( y /y 2 , y = p 3 /x = f 1 ( x ) 22. y = f 1 ( x ), x = f ( y ( 2 y, y 0 y 2 ,y> 0 , y = f 1 ( x ( x/ 2 ,x 0 x, x > 0 23. y = f 1 ( x ) = f ( y ( 5 / 2 y, y < 2 1 /y, y 2 ,y = f 1 ( x ( 5 / 2 x, x > 1 / 2 1 /x, 0 <x 1 / 2 24. y = p 1 ( x ), x = p ( y y 3 3 y 2 +3 y 1=( y 1) 3 , y = x 1 / 3 +1= p 1 ( x ) 25. y = f 1 ( x ), x = f ( y )=( y +2) 4 for y 0, y = f 1 ( x x 1 / 4 2 for x 16 26. y = f 1 (
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Calculus: Early Transcendentals, by Anton, 7th Edition,ch04...

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