Explanation Given information: The function is as, y = x 3 + x 2 − x ...... (1) Calculation: The given function is polynomial of degree 3 that is defined for all real value of x . To obtain the maximum and minimum value of the function, differentiate the function y = x 3 + x 2 − x with respect to x . d y d x = d d x ( x 3 + x 2 − x ) d y d x = 3 x 2 + 2 x − 1 To obtain the critical point of the function, equate the derivative of the function with 0. 3 x 2 + 2 x − 1 = 0 3 x 2 + 3 x − x − 1 = 0 3 x ( x + 1 ) − 1 ( x + 1 ) = 0 ( x + 1 ) ( 3 x − 1 ) = 0 Simplify the above equation to obtain the value of x . ( x + 1 ) ( 3 x − 1 ) = 0 x + 1 = 0 or 3 x − 1 = 0 x = − 1 or x = 1 3 Therefore, the critical points are − 1 and 1 3 . To obtain the maximum and minimum value of the function y = x 3 + x 2 − x on the interval [ − 2 , 2 ] , determine the value of the function on the endpoints of the interval and at the critical points that lies within the interval. The critical points − 1 and 1 3 lie on the interval [ − 2 , 2 ] . Substitute − 1 for x in the equation (1) to obtain the value of function at x = − 1 . y = ( − 1 ) 3 + ( − 1 ) 2 − ( − 1 ) = − 1 + 1 + 1 = 1 The value of function at x = − 1 is 1

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 4.2, Problem 30E

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Chapter 4.2, Problem 29E

Chapter 4.2, Problem 31E

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Chapter 4 Solutions

Calculus - Standalone book

Ch. 4.1 - Prob. 1PQ Ch. 4.1 - Prob. 2PQ Ch. 4.1 - Prob. 3PQ Ch. 4.1 - Prob. 4PQ 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 Ch. 4.1 - Prob. 6E

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The minimum and maximum value of the function y = x 3 + x 2 − x on the interval [ − 2 , 2 ] .

Solution Summary: The author explains that the given function is a polynomial of degree 3 that is defined for all real value of x.

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To find:

The minimum and maximum value of the function y=x3+x2−x on the interval [−2,2].

Want to see the full answer?

Chapter 4 Solutions

The minimum and maximum value of the function y = x 3 + x 2 − x on the interval [ − 2 , 2 ] .

Solution Summary: The author explains that the given function is a polynomial of degree 3 that is defined for all real value of x.

Concept explainers

Videos

To find: The minimum and maximum value of the function y=x3+x2−x on the interval [−2,2].

Want to see the full answer?

Chapter 4 Solutions

To find:

The minimum and maximum value of the function y=x3+x2−x on the interval [−2,2].