Explanation Given: The function is f ( x ) = − 2 x 3 − 1 2 x 2 − x − 3 . Calculation: Solve the given function: f ( x ) = − 2 x 3 − 1 2 x 2 − x − 3 Differentiate the above function: f ′ ( x ) = − 6 x 2 − x − 1 Substitute 0 for f ' ( x ) : − 6 x 2 − x − 1 = 0 Since the above function will have no real roots, it will have no local extrema points. Differentiate the above function again: f ″ ( x ) = − 12 x − 1 Substitute 0 for f " ( x ) : − 12 x − 1 = 0 x = − 1 12 The point of inflection of the graph is at x = − 1 12 . Let choose a point − 1 less than − 1 12 and 0 greater than − 1 12 . Calculate the value of f ″ ( 1 ) : f ″ ( 1 ) = − 12 ( 1 ) − 1 = − 13 Calculate the value of f ″ ( 0 ) : f ″ ( 0 ) = − 12 ( 0 ) − 1 = − 1 Since, f ″ ( 0 ) < 0 and f ″ ( 1 ) < 0 , the function is decreasing on ( − ∞ , ∞ ) . Now, we have to determine concavity of the graph by solving f " ( x ) > 0 a n d f " ( x ) < 0

Mathematics with Applications, Books a la Carte, and MyLab Math with Pearson eText - Title-Specific Access Card Package (12th Edition)

12th Edition

ISBN: 9780134862651

Author: Lial

Publisher: PEARSON

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Chapter 12, Problem 21RE

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To calculate: The graph of the functions and list the location of local extremum and point of inflection, the intervals where the function is increasing and decreasing and the interval where it is concave upward and concave downward.

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Chapter 12, Problem 20RE

Chapter 12, Problem 22RE

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Chapter 12 Solutions

Mathematics with Applications, Books a la Carte, and MyLab Math with Pearson eText - Title-Specific Access Card Package (12th Edition)

Ch. 12.1 - Checkpoint 1 For what values of x is the function...Ch. 12.1 - Checkpoint 2 Find all intervals on which is...Ch. 12.1 - Checkpoint 3 Identity the x-values of all points...Ch. 12.1 - Checkpoint 4 Find the critical numbers for each of...Ch. 12.1 - Prob. 5CP Ch. 12.1 - Prob. 6CP Ch. 12.1 - Checkpoint 7 Find the locations of the local...Ch. 12.1 - Prob. 8CP Ch. 12.1 - Checkpoint 9 If a sales function is given by...Ch. 12.1 - Prob. 1E

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