Calculus

# Calculus - Chapter 6 Exponentials, Logarithms and other...

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Chapter 6 Exponentials, Logarithms and other Transcendental Functions 184 Chapter 6 Section 6.1 5. 4 1 1 ln 4 dt t = 7. 8.2 1 1 ln8.2 dt t = 9. 11 (ln 4 ) 4 4 x xx =⋅ = 11. [] 1 ln(cos ) ( sin ) tan cos x x =− = 13. 1 ln ln 1 ln xx x x x x =⋅+ =+ 15. 2 22 12 ln( 2) 2 x  += =  ++ 17. 1/2 ln 2 3ln2 ln2 3ln2 1 ln 2 3ln 2 2 7 ln 2 2 + = 19. 21 / 2 2ln3 ln9 ln 3 2ln3 ln3 ln3 1 2ln3 2ln3 2 1 ln 3 2 −+ = + =−+ = 21. Use 2 1, ux then 2 du x dx = . 2 2 1 ln ln 1 x dx du u x uc xc = + + ∫∫ 23. Use u = cos x , then du = –sin xdx . sin sin cos cos 1 ln ln cos dx dx du u + + 25. Use 2 ux x =+− , then (2 2) 2( 1) du x dx x dx . 2 2( 2 2 1 ln 2 1 ln 2 1 2 x x dx dx du u c + + = +− = + 27. Use u = ln x , then 1 du dx x = . 1 ln ln 1 ln ln ln dx dx x x du u = = 29. Use ln 1 , then 1 du dx x = . 2 2 2 3 3 (ln 1 (ln 1 3 1 (ln 3 x dx x dx ud u + = + 31. 2 2 0 0 1 ln 1 ln 3 1 dx x x = +  = + 33. 2 1 1 3 3 0 0 1 ln 4 3 4 1 (ln3 ln 4) 3 13 ln 34 x dx x x = 35. 1 / 2 2 2 2 ln 1 ln( 1 ln( 2 2 2 1 1 dd dx dx d x dx x x x x +  = + = +

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Chapter 6 Exponentials, Logarithms and other Transcendental Functions 185 37. 4 45 5 4 5 ln ln ln( 1) 1 ln ln( 1 dx d xx dx dx x dd dx dx x x x  =− +   +   +   + 39. 2 11 1 1 1 2 2ln ln ln dx dx dx x x == ∫∫ Use ln ux = , then 1 du dx x = . 1 1 1 22 ln 1 ln 2 1 ln ln 2 dx du xu x uc xc = =+ 41. y 1 x 0 1 x = 2 y = ln( x – 2) Increasing and concave down on (2, ). 43. y 1 x 0 1 y = ln( x 2 + 1) Increasing on (0, ), decreasing on ( , 0). Concave up on ( 1,1); concave down on ( , 1) (1, ). ∞− −∞ − ∪ 45. y 1 x 0 1 y = x ln x Decreasing on 0, , increasing on , . Concave up on (0, ). ee    47. We will make use of the Reciprocal Property, 1 ln ln b b , in our proof. To prove this property, notice that ln ln ln ln b bb b =⋅ And since ln ln1 0 b b we must have that 1 ln ln b b (since the additive inverse of a real number is unique and since y = ln x is one-to-one). Therefore, 1 ln ln 1 ln ln ln ln a a a b ab as desired. 49. 1 () l n l n sx x x x x 2 1 l n 2 2l n (1 2 ln ) x x x x x =− − + The critical values are 1/2 0a n d12 l n 0 1 ln 2 x xe = = The function is undefined for x = 0 and the derivative changes from + to – at = . Therefore, there is a maximum at = .
Chapter 6 Exponentials, Logarithms and other Transcendental Functions 186 51. 2 2 3 () ln ln 1 ( ) , positive for . (ln ) 2l n ( ) , negative for . (ln ) x fx x x x e x x xe xx = => ′′ As x increases, the number of primes less than x increases, but at a decreasing rate. Section 6.2 5. 2 () 3 0 x =≥ for all x , so f is always increasing, hence one-to-one. 7. 2 2 , x =− which is positive for 2 3 x > and negative for 2 . 3 x < Hence f has a local maximum at 2 3 x and a local minimum at 2 , 3 x = so is not one-to-one. 9. The function is not one-to-one since it’s periodic. 11. 1 0 x for all x > 0, that is, for all x in the domain of f . So f is always increasing, hence one- to-one.

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## This homework help was uploaded on 04/19/2008 for the course MATH 105 taught by Professor Schroder during the Spring '08 term at Millersville.

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Calculus - Chapter 6 Exponentials, Logarithms and other...

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