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361_James Stewart Calculus 5 Edition Answers

# 361_James Stewart Calculus 5 Edition Answers - Stewart...

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h ( x )= x 2 ± 1 ( ) 3 ² h / ( x )=6 x x 2 ± 1 ( ) 2 ³ 0 ´ x >0 x µ 1 h 0, ( ) ± ¶ ,0 ( ) h (0)= ± 1 h / / ( x )=6 x 2 ± 1 ( ) 2 +24 x 2 x 2 ± 1 ( ) =6 x 2 ± 1 ( ) 5 x 2 ± 1 ( ) · 1 · 1 5 R x 2 5 x 2 ± 1 h / / x ( ) x < ± 1 + + + ± 1< x < ± 1 5 ± + ± ± 1 5 < x < 1 5 ± ± + 1 5 < x <1 ± + ± x >1 + + + h ± ¶ , ± 1 ( ) ± 1 5 , 1 5 1, ( ) ± 1, ± 1 5 1 5 ,1 · 1,0 ( ) · 1 5 , ± 64 125 A ( x )= x x +3 ² A / ( x )= x ¸ 1 2 ( x +3) ± 1/2 + x +3 ¸ 1= x 2 x +3 + x +3 = x +2( x +3) 2 x +3 = 3 x +6 2 x +3 A ± 3, ) A / ( x )>0 x > ± 2 A / ( x )<0 ± 3< x < ± 2 A 38. (a) ( ), so is increasing on and decreasing on . (b) is a local minimum value. (c) . The roots and divide into five intervals. Interval Concavity upward downward upward downward upward From the table, we see that
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