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# 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 ( x ) = 5 x 4 0 on ( −∞ , ) and f ( 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 ( x ) = 9 x 2 0 on ( −∞ , ) and f ( x ) = 0 only at x = 0; 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) 1 / 3 f 1 ( x ) = 1 3 ( x 1) 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 0 on ( −∞ , ) 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 0 on ( −∞ , ) and f ( x ) = 0 only at x = 1; f is increasing. Therefore, f is one-to-one. f ( y ) = x ( y + 1) 3 + 2 = x ( y + 1) 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 + 1) dom f 1 = ( −∞ , ) 13. f ( x ) = 3 5 x 2 / 5 > 0 for all x = 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 = 2 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)
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 0 on [ π/ 2 , π/ 2]; and f ( x ) = 0 only at x = π/ 2 , π/ 2. Therefore f is increasing on [ π/ 2 , π/ 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[ π/ 2 , π/ 2]. 18. f is not one-to-one on ( π/ 2 , π/ 2). For example, f ( π/ 4) = f ( π/ 4) = 2 2 . 19. f ( x ) = 1 x 2 < 0 for all x = 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 = 0 20. f 1 ( x ) = 1 1 x dom f 1 : x = 0 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 = 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 1 / 3 f 1 ( x ) = 1 x 1 1 / 3 dom f 1 : x = 0 24. f 1 ( x ) = x 1 + x dom f 1 : x = 1 25. f ( x ) = 1 ( x + 1) 2 < 0 for all x = 1; f is decreasing on ( −∞ , 1) ( 1 , ) f ( y ) = x y + 2 y + 1 = 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 = 1 26. not one-to-one; e.g. f (1) = f ( 3)

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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-07
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