Explanation Given information: The given function is, g ( x ) = x 2 + 2 x 2 3 (1) Theorem used: If f has a local maximum or local minimum value at c and f ′ ( c ) exists, then f ′ ( c ) = 0 . Calculation: The given function is continuous on the interval [ − 2 , 2 ] and differentiable on ( − 2 , 2 ) . First find the critical points on the graph of given function. Differentiate the given function to obtain the first derivative of g ( x ) , g ′ ( x ) = d d x ( 1 x + 1 ) = − 1 ( x + 1 ) 2 The above expression can never be equal to zero. g ′ ( x ) is not defined at x = 0 , so 0 is the critical point. And, the ends point are [ − 2 , 2 ]

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ISBN: 9781305041264

Author: Tan

Publisher: CENGAGE C

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Chapter 4.4, Problem 32E

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To find: The absolute maximum and minimum for the function g(x)=x2+2x23 on [−2,2] .

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Chapter 4.4, Problem 31E

Chapter 4.4, Problem 33E

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Ch. 4.1 - Explain each of the following: a. f is increasing...Ch. 4.1 - Prob. 2CQ Ch. 4.1 - Prob. 3CQ Ch. 4.1 - Prob. 4CQ Ch. 4.1 - Prob. 5CQ Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E

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The absolute maximum and minimum for the function g ( x ) = x 2 + 2 x 2 3 on [ − 2 , 2 ] .

Solution Summary: The author explains the absolute maximum and minimum values for the function g(x).

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To find: The absolute maximum and minimum for the function g(x)=x2+2x23 on [−2,2] .

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Chapter 4 Solutions