Explanation Given: The function is f ( x ) = x 3 − 12 x 2 + 45 x . Formula used: To find the critical points substitute f ′ ( x ) = 0 . Second derivative test: If f ′ ( c ) = 0 and f ″ ( c ) exists, then f ( c ) is a local maximum if f ″ ( c ) < 0 . f ( c ) is a local minimum if f ″ ( c ) > 0 . The test fails if f ″ ( c ) = 0 Calculation: The function is f ( x ) = x 3 − 12 x 2 + 45 x . We have to find the critical points and apply the second derivative test. First, find the derivative. f ′ ( x ) = 3 x 2 − 24 x + 45 Now, find the second derivative. f ″ ( x ) = 6 x − 24 Now, find the critical points by substitute f ′ ( x ) = 0 . 3 x 2 − 24 x + 45 = 0 x 2 − 8 x + 15 = 0 x 2 − 5 x − 3 x + 15 = 0 x ( x − 5 ) − 3 ( x − 5 ) = 0 ( x − 5 ) ( x − 3 ) = 0 That is, ( x − 5 ) = 0 or ( x − 3 ) = 0 x = 5 or x = 3 Thus, the critical points are 3 and 5

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 4.4, Problem 23E

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Find the critical points. Also, apply the second derivative test (or state that it fails).

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Chapter 4.4, Problem 22E

Chapter 4.4, Problem 24E

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Chapter 4 Solutions

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Find the critical points. Also, apply the second derivative test (or state that it fails).

Solution Summary: The author explains how to determine critical points and apply the second derivative test.

Find the critical points. Also, apply the second derivative test (or state that it fails).

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Chapter 4 Solutions