Explanation Given: The function y = x 4 − 3 x 3 + 4 x Key concepts used: Critical values of a function can be found by solving the equation f ' = 0 . The sign of f ' is positive when graph is increasing. The sign of f ' is negative when graph is decreasing. The sign of f ' ' is positive when graph is concave up. The sign of f ' ' is negative when graph is concave down. Maximum of a function occurs when derivative changes its sign from positive to negative. Minimum of a function occurs when derivative changes its sign from negative to positive. Inflection point occurs when double derivative changes its sign, provided the original function is defined at that point. Calculation: Let us first find critical values of the given function. For that, first we find the derivative of the function and set it equal to zero: ⇒ y = x 4 − 3 x 3 + 4 x ⇒ y ' = 4 x 3 − 9 x 2 + 4 ⇒ y ' = 4 x 3 − 9 x 2 + 4 = 0 ⇒ x = − 0.59 , 2 , 0.84 Thus, the critical values of the given function are x = − 0.59 , 2 , 0.84 Function is increasing when derivative is positive: ⇒ y ' > 0 ⇒ 4 x 3 − 9 x 2 + 4 > 0 ⇒ x ∈ ( − 0.59 , 0.84 ) ∪ ( 2 , ∞ ) Function is decreasing when derivative is negative: ⇒ y ' < 0 ⇒ 4 x 3 − 9 x 2 + 4 < 0 ⇒ x ∈ ( − ∞ , − 0

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 4.5, Problem 23E

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Chapter 4.5, Problem 22E

Chapter 4.5, Problem 24E

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Chapter 4 Solutions

CALCULUS (CLOTH)

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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To graph: The given function.

Solution Summary: The author explains how critical values of a function can be found by solving the equation f'=0.

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Chapter 4 Solutions

To graph:

The given function.