Calculus: One and Several Variables

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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-07 JWDD027-Salas-v1 November 25, 2006 15:41 SECTION 7.1 341 CHAPTER 7 SECTION 7.1 1. Suppose f ( x 1 )= f ( x 2 ) x 1 ± = x 2 . Then 5 x 1 +3=5 x 2 +3 x 1 = x 2 ; f is one-to-one f ( y x 5 y +3= x 5 y = x 3 y = 1 5 ( x 3) f 1 ( x 1 5 ( x 3) dom f 1 =( −∞ , ) 2. f 1 ( x 1 3 ( x 5) dom f 1 −∞ , ) 3. f is not one-to-one; e.g. f (1) = f ( 1) 4. f 1 ( x x 1 / 5 ; dom f 1 −∞ , ) 5. f ² ( x )=5 x 4 0on( −∞ , ) and f ² ( x ) = 0 only at x =0; f is increasing. Therefore, f is one-to-one. f ( y x y 5 +1= x y 5 = x 1 y x 1) 1 / 5 f 1 ( x )=( x 1) 1 / 5 dom f 1 −∞ , ) 6. not one-to-one; e.g. f (0) = f (3) 7. f ² ( x )=9 x 2 0o n( −∞ , ) and f ² ( x ) = 0 only at x f is increasing. Therefore, f is one-to-one. f ( y x 1+3 y 3 = x y 3 = 1 3 ( x 1) y = ± 1 3 ( x 1) ² 1 / 3 f 1 ( x ± 1 3 ( x 1) ² 1 / 3 dom f 1 −∞ , ) 8. f 1 ( x x +1) 1 / 3 dom f 1 −∞ , )
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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-07 JWDD027-Salas-v1 November 25, 2006 15:41 342 SECTION 7.1 9. f ± ( x ) = 3(1 x ) 2 0on( −∞ , ) and f ± ( x ) = 0 only at x =1; f is increasing. Therefore, f is one-to-one. f ( y )= x (1 y ) 3 = x 1 y = x 1 / 3 y =1 x 1 / 3 f 1 ( x )=1 x 1 / 3 dom f 1 =( −∞ , ) 10. not one-to-one; e.g. f (0) = f (2) . 11. f ± ( x )=3( x +1) 2 0o n( −∞ , ) and f ± ( x ) = 0 only at x = 1; f is increasing. Therefore, f is one-to-one. f ( y x ( y 3 +2= x ( y 3 = x 2 y +1=( x 2) 1 / 3 y x 2) 1 / 3 1 f 1 ( x )=( x 2) 1 / 3 1 dom f 1 −∞ , ) 12. f 1 ( x 1 4 ( x 1 / 3 dom f 1 −∞ , ) 13. f ± ( x 3 5 x 2 / 5 > 0 for all x ² =0; f is increasing on ( −∞ , ) f ( y x y 3 / 5 = x y = x 5 / 3 f 1 ( x x 5 / 3 dom f 1 −∞ , ) 14. f 1 ( x )=(1 x ) 3 +2 dom f 1 −∞ , ) 15. f ± ( x 3(2 3 x ) 2 0 for all x and f ± ( x ) = 0 only at x =2 / 3; f is decreasing f ( y x (2 3 y ) 3 = x 2 3 y = x 1 / 3 3 y x 1 / 3 y = 1 3 (2 x 1 / 3 ) f 1 ( x 1 3 (2 x 1 / 3 ) dom f 1 −∞ , ) 16. not one-to-one; e.g. f (1) = f ( 1)
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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-07 JWDD027-Salas-v1 November 25, 2006 15:41 SECTION 7.1 343 17. f ± ( x ) = cos x 0o n[ π/ 2 ,π/ 2]; and f ± ( x ) = 0 only at x = 2 2. Therefore f is increasing on [ 2 2] and f has an inverse. The inverse is denoted by arcsin x ; this function will be studied in Section 7.7. The domain of arcsin x is [ 1 , 1] = range of sin x on[ 2 2]. 18. f is not one-to-one on ( 2 2). For example, f ( 4) = f ( 4) = 2 2 . 19. f ± ( x )= 1 x 2 < 0 for all x ² =0; f is decreasing on ( −∞ , 0) (0 , ) f ( y x = 1 y = x = y = 1 x f 1 ( x 1 x dom f 1 : x ² =0 20. f 1 ( x )=1 1 x dom f 1 : x ² 21. f is not one-to-one; e.g. f ( 1 2 ) = f (2) 22. not one-to-one; e.g. f (1) = f (2) 23. f ± ( x 3 x 2 ( x 3 +1) 2 0 for all x ² = 1; f is decreasing on ( −∞ , 1) ( 1 , ) f ( y x 1 y 3 +1 = x y 3 +1= 1 x y 3 = 1 x 1 y = ± 1 x 1 ² 1 / 3 f 1 ( x ± 1 x 1 ² 1 / 3 dom f 1 : x ² 24. f 1 ( x x 1+ x dom f 1 : x ² = 1 25. f ± ( x 1 ( x 2 < 0 for all x ² = 1; f is decreasing on ( −∞ , 1) ( 1 , ) f ( y x y +2 y = x y +2= xy + x y (1 x x 2 y = x 2 1 x f 1 ( x x 2 1 x dom f 1 : x ² =1 26. not one-to-one; e.g. f (1) = f ( 3)
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