377_James Stewart Calculus 5 Edition Answers

377_James Stewart Calculus 5 Edition Answers - Stewart...

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c = x 1 < x 2 x 1 < c < x 2 f x 1 ,c c,x 2 f ( x 1 )< f ( c )< f ( x 2 ) f ( x 1 )< f ( x 2 ) x 1 x 2 I x 1 < x 2 f I f I f / / >0 I f g I f / / >0 g / / >0 I f + g ( ) / / = f / / + g / / >0 I ± f + g I f I f >0 f / / >0 I g x ( ) =[ f x ( ) ] 2 ± g / =2 ff / ± g / / =2 f / f / +2 ff / / =2 f / ( ) 2 +2 ff / / >0 ± g I f g I f / / g / / 0 f >0 f / ² 0 f / / >0 g >0 g / ² 0 g / / >0 I ( fg ) / = f / g + fg / ± ( fg ) / / = f / / g +2 f / g / + fg / / ² f / / g + fg / / >0 I ± fg I f g f / ³ 0 g / ³ 0 I 2 f / g / ² 0 I ( fg ) / / = f / / g +2 f / g / + fg / / ² f / / g + fg / / >0 I ± fg I f g f / ² 0 g / ³ 0 I 2 f / g / ³ 0 I I = 0, ´ ( ) f ( x )= x 3 g ( x )=1/ x ( fg )( x )= x 2 ( fg ) / ( x )=2 x ( fg ) / / ( x )=2>0 I fg I I = 0, ´ ( ) f ( x )=4 x x g ( x )=1/ x ( fg )( x )=4 x ( fg ) / ( x )=2/ x ( fg ) / / ( x )= µ 1/ x 3 <0 I fg I I = 0, ´ ( ) f ( x )= x 2 g ( x )=1/ x ( fg )( x )= x fg I f g µ ´ , ´ ( ) f / / >0 g / / >0 µ ´ , ´ ( ) h ( x )= f ( g ( x )) ± h / ( x )= f / ( g ( x )) g / ( x ) ± Case 4 . Same proof as in Case 3. Case 5 . By Cases 3 and 4, is increasing on and on , so . In all cases, we have shown that . Since , were any numbers in with , we have shown that is increasing on . 66. (a) We will make use of the converse of the Concavity Test (along with the stated assumptions); that is, if is concave upward on , then on . If and are CU on , then and on , so on is CU on . (b) Since is positive and CU on , and on . So is CU on . 67. (a) Since and are positive, increasing, and CU on with
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This note was uploaded on 08/02/2011 for the course MATH 1B taught by Professor Reshetiken during the Spring '08 term at University of California, Berkeley.

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