Explanation Definitions used: Concavity and inflection point: Let f be differentiable on an open interval I . If f ′ is increasing on I , then f is concave up on I . If f ′ is decreasing on I , then f is concave down on I . If f is continuous at c and f changes concavity at c (from up to down, or vice versa), then f has an inflection point at c . Theorem used: Test for Concavity: Suppose that f ′ ′ exists on an open interval I . (1) If f ′ ′ > 0 on I , the f is concave up on I. (2) If f ′ ′ < 0 on I , then f is concave down on I . (3) If c is a point of I at which f ′ ′ changes sign at c (from positive to negative, or vice versa), then f has an inflection point at c . Calculation: The given function is g ( t ) = 3 t 5 − 30 t 4 + 80 t 3 + 100 . Differentiate g ( t ) = 3 t 5 − 30 t 4 + 80 t 3 + 100 with respect to t on both sides, g ′ ( t ) = d d t ( 3 t 5 − 30 t 4 + 80 t 3 + 100 ) = 3 ( 5 t 4 ) − 30 ( 4 t 3 ) + 80 ( 3 t 2 ) g ′ ( t ) = 15 t 4 − 120 t 3 + 240 t 2 Obtain the second derivative of the function g ( t ) = 3 t 5 − 30 t 4 + 80 t 3 + 100 . Differentiate g ′ ( t ) = 15 t 4 − 120 t 3 + 240 t 2 with respect to t on both sides, g ″ ( t ) = d d t ( 15 t 4 − 120 t 3 + 240 t 2 ) = 60 t 3 − 360 t 2 + 480 t = 60 t ( t 2 − 6 t + 8 ) = 60 t ( t − 2 ) ( t − 4 ) Set the above derivative to zero and obtain the candidates of inflection point. 60 t ( t − 2 ) ( t − 4 ) = 0 60 t = 0 ; ( t − 2 ) = 0 ; ( t − 4 ) = 0 t = 0 ; t = 2 ; t = 4 The inflection points occurs at t = 0 , t = 2 and t = 4 . In the interval notation, ( − ∞ , 0 ) ∪ ( 0 , 2 ) ∪ ( 2 , 4 ) ∪ ( 4 , ∞ ) . Find the nature of function in these intervals ( − ∞ , 0 ) , ( 0 , 2 ) , ( 2 , 4 ) , ( 4 , ∞ ) as follows

Single Variable Calculus: Early Transcendentals, Books a la Carte, and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition)

3rd Edition

ISBN: 9780134996103

Author: William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz

Publisher: PEARSON

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Chapter 4.3, Problem 75E

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To find: The intervals on which the function g(t)=3t5−30t4+80t3+100 is concave up or concave down and also identify the inflection points.

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Chapter 4.3, Problem 74E

Chapter 4.3, Problem 76E

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Single Variable Calculus: Early Transcendentals, Books a la Carte, and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition)

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The intervals on which the function g ( t ) = 3 t 5 − 30 t 4 + 80 t 3 + 100 is concave up or concave down and also identify the inflection points.

Solution Summary: The author explains how the function g(t) is concave up, and the inflection points.

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To find: The intervals on which the function g(t)=3t5−30t4+80t3+100 is concave up or concave down and also identify the inflection points.

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Chapter 4 Solutions