Explanation Given information: We have g ( 0 ) = 0 . The figure gives the derivative of g ' . Formula used: Fundamental theorem of calculus : If f is continuous on the closed interval [ a , b ] and F is the indefinite integral of f on [ a , b ] , then ∫ a b f ( x ) d x = F ( b ) − F ( a ) Calculation: From the given graph, we can easily observe that g ' ( x ) = { x , x ∈ [ 0 , 1 ] 1 , x ∈ [ 1 , 3 ] 4 − x , x ∈ [ 3 , 5 ] x − 6 , x ∈ [ 5 , 6 ] By the fundamental theorem, we have g ( x ) = g ( 0 ) + ∫ 0 x g ' ( t ) d t Since, g ( 0 ) = 0 , g ( x ) = ∫ 0 x g ' ( t ) d t Local maxima and minima must satisfy the condition g ' ( x ) = 0 , which is possible for x = 0 and x = 4

6th Edition

ISBN: 9781119399353

Author: DA

Publisher: JOHN WILEY+SONS,INC.-CONSIGNMENT

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Chapter 6.1, Problem 11P

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(x,y)− coordinates of local maxima and minima and to graph gif g(0)=0.

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Chapter 6.1, Problem 10P

Chapter 6.1, Problem 12P

Chapter 6 Solutions

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( x , y ) − coordinates of local maxima and minima and to graph gif g ( 0 ) = 0 .

Solution Summary: The author calculates the derivative of g'. If f is continuous on the closed interval [a,b] and F is the indefinite integral, then local maxima and minima must satisfy

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To find:

(x,y)− coordinates of local maxima and minima and to graph gif g(0)=0.

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Chapter 6 Solutions

( x , y ) − coordinates of local maxima and minima and to graph gif g ( 0 ) = 0 .

Solution Summary: The author calculates the derivative of g'. If f is continuous on the closed interval [a,b] and F is the indefinite integral, then local maxima and minima must satisfy

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To find: (x,y)− coordinates of local maxima and minima and to graph gif g(0)=0.

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Chapter 6 Solutions

To find:

(x,y)− coordinates of local maxima and minima and to graph gif g(0)=0.