Chapter 3 - SECTION 3.1 (PAGE 167) R. A. ADAMS: CALCULUS...

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SECTION 3.1 (PAGE 167) R. A. ADAMS: CALCULUS CHAPTER 3. TRANSCENDENTAL FUNC- TIONS Section 3.1 Inverse Functions (page 167) 1. f ( x ) = x 1 f ( x 1 ) = f ( x 2 ) x 1 1 = x 2 1 x 1 = x 2 . Thus f is one-to-one. Let y = f 1 ( x ) . Then x = f ( y ) = y 1 and y = x + 1. Thus f 1 ( x ) = x + 1. D ( f ) = D ( f 1 ) = = R ( f ) = R ( f 1 ) . 2. f ( x ) = 2 x 1. If f ( x 1 ) = f ( x 2 ) , then 2 x 1 1 = 2 x 2 1. Thus 2 ( x 1 x 2 ) = 0 and x 1 = x 2 . Hence, f is one-to- one. Let y = f 1 ( x ) . Thus x = f ( y ) = 2 y 1, so y = 1 2 ( x + 1 ) . Thus f 1 ( x ) = 1 2 ( x + 1 ) . D ( f ) = R ( f 1 ) = ( −∞ , ) . R ( f ) = D ( f 1 ) = ( −∞ , ) . 3. f ( x ) = x 1 f ( x 1 ) = f ( x 2 ) ± x 1 1 = ± x 2 1 ,( x 1 , x 2 1 ) x 1 1 = x 2 1 = 0 x 1 = x 2 Thus f is one-to-one. Let y = f 1 ( x ) . Then x = f ( y ) = y 1, and y = 1 + x 2 . Thus f 1 ( x ) = 1 + x 2 , ( x 0 ) . D ( f ) = R ( f 1 ) = [1 , ) , R ( f ) = D ( f 1 ) = [0 , ) . 4. f ( x ) =− x 1 for x 1. If f ( x 1 ) = f ( x 2 ) , then x 1 1 x 2 1 and x 1 1 = x 2 1. Thus x 1 = x 2 and f is one-to-one. Let y = f 1 ( x ) . Then x = f ( y ) y 1so x 2 = y 1 and y = x 2 + 1. Thus, f 1 ( x ) = x 2 + 1. D ( f ) = R ( f 1 ) = [1 , ) . R ( f ) = D ( f 1 ) = ( −∞ , 0]. 5. f ( x ) = x 3 f ( x 1 ) = f ( x 2 ) x 3 1 = x 3 2 ( x 1 x 2 )( x 2 1 + x 1 x 2 + x 2 2 ) = 0 x 1 = x 2 Thus f is one-to-one. Let y = f 1 ( x ) . Then x = f ( y ) = y 3 so y = x 1 / 3 . Thus f 1 ( x ) = x 1 / 3 . D ( f ) = D ( f 1 ) = = R ( f ) = R ( f 1 ) . 6. f ( x ) = 1 + 3 x .I f f ( x 1 ) = f ( x 2 ) , then 1 + 3 x 1 = 1 + 3 x 2 so x 1 = x 2 . Thus, f is one-to- one. Let y = f 1 ( x ) so that x = f ( y ) = 1 + 3 y . Thus y = ( x 1 ) 3 and f 1 ( x ) = ( x 1 ) 3 . D ( f ) = R ( f 1 ) = ( −∞ , ) . R ( f ) = D ( f 1 ) = ( −∞ , ) . 7. f ( x ) = x 2 x 0 ) f ( x 1 ) = f ( x 2 ) x 2 1 = x 2 2 x 1 0 , x 2 0 ) x 1 = x 2 Thus f is one-to-one. Let y = f 1 ( x ) . Then x = f ( y ) = y 2 ( y 0 ) . therefore y x and f 1 ( x ) x . D ( f ) = ( −∞ , 0] = R ( f 1 ) , D ( f 1 ) = [0 , ) = R ( f ) . 8. f ( x ) = ( 1 2 x ) 3 f f ( x 1 ) = f ( x 2 ) , then ( 1 2 x 1 ) 3 = ( 1 2 x 2 ) 3 and x 1 = x 2 . Thus, f is one-to- one. Let y = f 1 ( x ) . Then x = f ( y ) = ( 1 2 y ) 3 so y = 1 2 ( 1 3 x ) . Thus, f 1 ( x ) = 1 2 ( 1 3 x ) . D ( f ) = R ( f 1 ) = ( −∞ , ) . R ( f ) = D ( f 1 ) = ( −∞ , ) . 9. f ( x ) = 1 x + 1 . D ( f ) ={ x : x ±=− 1 }= R ( f 1 ) . f ( x 1 ) = f ( x 2 ) 1 x 1 + 1 = 1 x 2 + 1 x 2 + 1 = x 1 + 1 x 2 = x 1 Thus f is one-to-one; Let y = f 1 ( x ) . Then x = f ( y ) = 1 y + 1 so y + 1 = 1 x and y = f 1 ( x ) = 1 x 1. D ( f 1 ) x : x ±= 0 R ( f ) . 10. f ( x ) = x 1 + x f f ( x 1 ) = f ( x 2 ) , then x 1 1 + x 1 = x 2 1 + x 2 . Hence x 1 ( 1 + x 2 ) = x 2 ( 1 + x 1 ) and, on simplification, x 1 = x 2 . Thus, f is one-to-one. Let y = f 1 ( x ) . Then x = f ( y ) = y 1 + y and x ( 1 + y ) = y . Thus y = x 1 x = f 1 ( x ) . D ( f ) = R ( f 1 ) = ( −∞ , 1 ) ( 1 , ) . R ( f ) = D ( f 1 ) = ( −∞ , 1 ) ( 1 , ) . 11. f ( x ) = 1 2 x 1 + x . D ( f ) x : x 1 R ( f 1 ) f ( x 1 ) = f ( x 2 ) 1 2 x 1 1 + x 1 = 1 2 x 2 1 + x 2 1 + x 2 2 x 1 2 x 1 x 2 = 1 + x 1 2 x 2 2 x 1 x 2 3 x 2 = 3 x 1 x 1 = x 2 Thus f is one-to-one. Let y = f 1 ( x ) . Then x = f ( y ) = 1 2 y 1 + y so x + xy = 1 2 y and f 1 ( x ) = y = 1 x 2 + x . D ( f 1 ) x : x 2 R ( f ) .
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Chapter 3 - SECTION 3.1 (PAGE 167) R. A. ADAMS: CALCULUS...

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