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MS-Sol-EE-C05

MS-Sol-EE-C05 - CHAPTER 5 LIMITS AND DERIVATIVES EXERCISE...

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CHAPTER 5 LIMITS AND DERIVATIVES EXERCISE 5.1 Section 5.1 Idea of functions (page 141) 1. domain: all real numbers x 2 0 x 2 + 2 2 range: [2, ) 2. domain: all real numbers e x 0 e x - 1 - 1 range: [ - 1, ) 3. We cannot take the logarithm of a negative number. x - 3 0 x 3 domain: [3, ) range: all real numbers 4. p ( x ) is undefined if its denominator = 0 x + 4 0 x - 4 domain: all real numbers except - 4 k 1 0 for all k range: all real numbers except 0 5. We cannot take the square root of a negative number. domain: [0, ) x 0 - x 0 range: ( -∞ , 0] 120

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C HAPTER 5 L IMITS AND D ERIVATIVES 6. G ( x ) is undefined if its denominator = 0 x - 1 0 x 1 domain: all real numbers except 1 G ( x ) = 1 1 - + x x = 1 + 1 2 - x 1 2 - x 0 for all x range: all real numbers except 1 7. We cannot take the square root of a negative number. ( x - 1) ( x - 5) 0 x 1 or x 5 domain: ( -∞ , 1] and [5, ) k 0 for all k range: [0, ) 8. T ( x ) is undefined if its denominator = 0 x - 6 0 x 6 We cannot take the square root of a negative number. 6 - x > 0 x < 6 domain: ( -∞ , 6) k > 0 for all k > 0 range: (0, ) 9. y = g ( h ( x )) = g ( x 2 + 2 x - 1) = ( x 2 + 2 x - 1) 3 10. y = g ( h ( x )) = g - + x x 3 3 = x x - + 3 3 121
S ECTION 5.1 I DEA OF F UNCTIONS 11. y = g ( h ( x )) = g - 2 2 x = 10 e 2 2 x - 12. y = g ( h ( x )) = g [ln (1 + x 3 )] = 2 ) 1 ( ln 3 + + x 13. g (3) = 3 2 = 9 f ( g (3)) = f (9) = 1 9 - = 2 2 14. h ( e 5 ) = ln e 5 = 5 f ( h ( e 5 )) = f (5) = 1 5 - = 2 15. g ( e ) = e 2 h ( g ( e )) = h ( e 2 ) = ln e 2 = 2 16. f (5) = 1 5 - = 2 g ( f (5)) = g (2) = 2 2 = 4 17. f ( g ( x )) = f ( x + 1) = ( x + 1) 2 domain: all real numbers k 2 0 for all k range: [0, ) g ( f ( x )) = g ( x 2 ) = x 2 + 1 domain: all real numbers x 2 0 for all x x 2 + 1 1 range: [1, ) 18. f ( g ( x )) = f (5 e x ) 122

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C HAPTER 5 L IMITS AND D ERIVATIVES = x e 5 domain: all real numbers e x > 0 for all x range: (0, ) g ( f ( x )) = g ( x ) = 5 e x We cannot take the square root of a negative number. x 0 domain: [0, ) x 0 e x 1 5 e x 5 range: [5, ) 19. f ( g ( x )) = f (ln x ) = 3 ln 1 - x It is undefined if its denominator = 0 ln x - 3 0 ln x 3 x e 3 domain: all reall numbers except e 3 k 1 0 for all k range: all real numbers except 0 g ( f ( x )) = g - 3 1 x = ln 3 1 - x We can only take the logarithm of a positive number. 3 1 - x > 0 x - 3 > 0 x > 3 domain: (3, ) range: all real numbers 123
C HAPTER 5 L IMITS AND D ERIVATIVES 20. f ( g ( x )) = f (ln x 2 ) = 4 ln 2 - x We cannot take the square root of a negative number. ln x 2 - 4 0 ln x 2 4 x 2 e 4 x - e 2 or x e 2 domain: ( -∞ , - e 2 ] and [ e 2 , ) k 0 for all k range: [0, ) g ( f ( x )) = g ( 4 - x ) = ln ( 4 - x ) 2 = ln ( x - 4) We can only take the logarithm of a positive number. x - 4 > 0 x > 4 domain: (4, ) range: all real numbers EXERCISE 5.2 Section 5.2 Limits of functions (page 147) 1. lim ( ) x c f x - = 3 lim ( ) x c f x + = 3 lim ( ) x c f x = 3 2.

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MS-Sol-EE-C05 - CHAPTER 5 LIMITS AND DERIVATIVES EXERCISE...

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