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345_James Stewart Calculus 5 Edition Answers

# 345_James Stewart Calculus 5 Edition Answers - Stewart...

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f ( x )=2 x ± 1 ± sin x f (0)= ± 1<0 f ( ± /2)= ± ± 2>0 f 2 x ± 1 ± 1 ( ) ² sin x sin x f x c 0, ± /2 ( ) f ( c )=0 a b a < b f ( a )= f ( b )=0 f a,b a,b ( ) r a,b ( ) f / ( r )=0 f / ( r )=2 ± cos r >0 cos r ³ 1 f ( x )= x 3 ± 15 x + c x ± 2,2 f a b ± 2,2 a < b f ( a )= f ( b )=0 f a,b a,b ( ) r a,b ( ) f / ( r )=0 f / ( r )=3 r 2 ± 15 r a,b ( ) ± 2,2 r <2 r 2 <4 3 r 2 ± 15<3 ² 4 ± 15= ± 3<0 f / ( r )=0 ± 2,2 ± 2,2 f ( x )= x 4 +4 x + c f ( x )=0 a b d a < b < d f ( a )= f ( b )= f ( d )=0 c 1 c 2 a < c 1 < b b < c 2 < d 0= f / c 1 ( ) = f / c 2 ( ) f / ( x )=0 0= f / ( x )=4 x 3 +4=4 x 3 +1 ( ) =4 x +1 ( ) x 2 ± x +1 ( ) x = ± 1 f ( x ) P ( x ) a 1 < a 2 < a 3 < a 4 P a 1 ( ) = P a 2 ( ) = P a 3 ( ) = P a 4 ( ) c 1 c 2 c 3 a 1 < c 1 < a 2 a 2 < c 2 < a 3 a 3 < c 3 < a 4 P / c 1 ( ) = P / c 2 ( ) = P / c 3 ( ) =0 ± P / ( x ) n n n =1 n P ( x ) n +1 P ( x ) n +1 a 1 < a 2 < a 3 < ² ² ² < a n +1 < a n +2 P a 1 ( ) = P a 2 ( ) = ² ² ² = P a n +2 ( ) =0 c 1 , ² ² ² ,c n +1 a 1 < c 1 < a 2 , ² ² ² , a n +1 < c n +1 < a n +2 P / c 1 ( ) = ² ² ² = P / c n +1 ( ) =0 n P / ( x ) n +1 P ( x ) n +1 18. Let . Then and . is the sum of the polynomial and the scalar multiple of the trigonometric function , so is continuous (and differentiable) for all . By the Intermediate Value Theorem, there is a number in such that . Thus, the given equation has at least one real root. If the equation has distinct real roots
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