Explanation Definition used: Critical point: A point whose x- coordinate is the critical number c such that either f ′ ( c ) = 0 or f ′ ( c ) does not exists and y -coordinate is f ( c ) . Increasing function: Let f be a function with some derivative defined on some interval. Then the function is said to be increasing if for any x in the interval, the inequality f ′ ( x ) > 0 holds. Decreasing function: Let f be a function with some derivative defined on some interval. Then the function is said to be decreasing if for any x in the interval, the inequality f ′ ( x ) < 0 holds. Calculation: The given function is f ( x ) = − x 3 + 2 x 2 + 15 x + 16 . Obtain the derivative as follows. f ′ ( x ) = − ( 3 x 2 ) + 2 ( 2 x ) + 15 ( 1 ) + 0 = − 3 x 2 + 4 x + 15 Set y ′ ( c ) = 0 and compute the value of c . − 3 c 2 + 4 c + 15 = 0 − 3 c 2 + 9 c − 5 c + 15 = 0 − 3 c ( c − 3 ) − 5 ( c − 3 ) = 0 ( − 3 c − 5 ) ( c − 3 ) = 0 That is, ( − 3 c − 5 ) ( c − 3 ) = 0 − 3 c − 5 = 0 or c − 3 = 0 c = − 5 3 or c = 3 Thus, the critical numbers is − 5 3 and 3 . These points determines the intervals, ( − ∞ , − 5 3 ) ∪ ( − 5 3 , 3 ) ∪ ( 3 , ∞ )

Finite Mathematics and Calculus with Applications

1st Edition

ISBN: 9781323188361

Author: Margaret Lial

Publisher: Pearson Education

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Chapter 13, Problem 19RE

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To find: The open intervals where f(x)=−x3+2x2+15x+16 increases or decreases.

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Chapter 13, Problem 18RE

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Finite Mathematics and Calculus with Applications

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The open intervals where f ( x ) = − x 3 + 2 x 2 + 15 x + 16 increases or decreases.

Solution Summary: The author explains that the function increases on the interval (-53, 3) and decreases if x holds the inequality.

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To find: The open intervals where f(x)=−x3+2x2+15x+16 increases or decreases.

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Chapter 13 Solutions