Calculus: Early Transcendentals - Instructor's Solution by James Stewart

Calculus: Early Transcendentals - Instructor's Solution by James Stewart

By James Stewart

In-Depth recommendations key to my prior add, Calculus: Early Transcendentals 5th variation. Explains the how to acquiring the solutions for all strange or even difficulties. the proper device for studying Calculus 1-3. take pleasure in.

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Extra info for Calculus: Early Transcendentals - Instructor's Solution Manual (5th Edition)

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The initial population is 100 , so the population y at t/3 time t is y=100 2 (c) t=20 . 20/3 y=100 2  10 , 159 x/3 (d) We graph y =100 2 1 and y =50 , 000 . 9 hours. 1 4 1 = g 2 8 (b) In t hours, there will be t/15 half life periods. The initial mass is 2 g, so the mass y at time t is 1 t/15 y=2 . 024 g (c) 4 days =4 24=96 hours. 7 hours 26. (a) Sixty hours represents 4 half life periods. 2 27. An exponential model is 7 Stewart Calculus ET 5e 0534393217;1. 154832569 10 and y(2010) 7417 million.

3 52. Let h(x)= x , g(x)=x 1 , and f (x)= x . Then ( f  g h)(x)= 4 3 x  1 =H(x) . 53. Let h(x)= x , g(x)=sec x , and f (x)=x . Then ( f  g h)(x)= ( sec x ) 4 =sec 4 ( x ) =H(x) . 54. (a) f (g(1))= f (6)=5 (b) g( f (1))=g(3)=2 (c) f ( f (1))= f (3)=4 (d) g(g(1))=g(6)=3 (e) (g f )(3)=g( f (3))=g(4)=1 (f) ( f  g)(6)= f (g(6))= f (3)=4 55. (a) g(2)=5 , because the point ( 2,5) is on the graph of g . Thus, f (g(2))= f (5)=4 , because the point ( 5,4 ) is on the graph of f . (b) g( f (0))=g(0)=3 13 Stewart Calculus ET 5e 0534393217;1.

The larger the base, the faster the function increases for x>0 , and the faster it approaches 0 as x   .  x 4. The graph of e  x x is the reflection of the graph of e about the y axis, and the graph of 8 is the x x x reflection of that of 8 about the y  axis. The graph of 8 increases more quickly than that of e for x>0 , and approaches 0 faster as x   . x x 5. The functions with bases greater than 1 ( 3 and 10 ) are increasing, while those with bases less x 1 x 1 1 x x than 1 and are decreasing.

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