Explanation Given: The provided function is, f ( x ) = − 4 x 3 − x 2 + 4 x + 5 Graph: The provided function is, f ( x ) = − 4 x 3 − x 2 + 4 x + 5 Now, First derivative of function f ( x ) = − 4 x 3 − x 2 + 4 x + 5 is, f ′ ( x ) = d d x ( − 4 x 3 − x 2 + 4 x + 5 ) = − 12 x 2 − 2 x + 4 Second derivative of function f ( x ) = − 4 x 3 − x 2 + 4 x + 5 is, f ″ ( x ) = d d x f ′ ( − 12 x 2 − 2 x + 4 ) = − 24 x − 2 Put f ′ ( x ) = 0 then, − 12 x 2 − 2 x + 4 = 0 − 2 ( 6 x 2 − x + 2 ) = 0 6 x 2 − 2 + 2 = 0 ( 2 x − 1 ) ( 3 x + 2 ) = 0 Further simplify then, ( 2 x − 1 ) = 0 2 x = 1 x = 1 2 And if, ( 3 x + 2 ) = 0 3 x = − 2 x = − 2 3 So, x = 1 2 and x = − 2 3 Put these critical numbers in equation f ″ ( x ) = − 24 x − 2 . f ″ ( x ) = − 24 x − 2 f ″ ( 1 2 ) = − 24 ( 1 2 ) − 2 = − 12 − 2 = − 14 < 0 So, there is a local maximum when x = 1 2 , Then, f ( x ) = − 4 x 3 − x 2 + 4 x + 5 f ( 1 2 ) = − 4 ( 1 2 ) 3 − ( 1 2 ) 2 + 4 × ( 1 2 ) + 5 = − 0.5 − 0.25 + 2 + 5 = 6.25 So, the value of f ( 1 2 ) = 6.25 . Then the point is ( 1 2 , 6.25 ) . Now, f ″ ( x ) = − 24 x − 2 f ″ ( − 2 3 ) = − 24 ( − 2 3 ) − 2 = 16 − 2 = 14 Now, f ″ ( − 2 3 ) > 0 So, there is a local minimum when x = − 2 3 , then, f ( x ) = − 4 x 3 − x 2 + 4 x + 5 f ( − 2 3 ) = − 4 ( − 2 3 ) 3 − ( − 2 3 ) 2 + 4 × ( − 2 3 ) + 5 = 4 × 8 27 − 4 9 − 8 3 + 5 = 3

Mathematics with Applications and Mylab Math with Pearson EText -- Title-Specific Access Card Package

12th Edition

ISBN: 9780134862668

Author: Lial, Margaret L.

Publisher: Pearson Education Canada

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Chapter 12, Problem 27RE

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Chapter 12 Solutions

Mathematics with Applications and Mylab Math with Pearson EText -- Title-Specific Access Card Package

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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To graph: The function f ( x ) = − 4 x 3 − x 2 + 4 x + 5 and then find the characteristics of the graph.

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To graph: The function f(x)=−4x3−x2+4x+5 and then find the characteristics of the graph.

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Chapter 12 Solutions