Explanation Given information:- The given derivative of y = x 3 − 48 / x 2 . Calculations:- The given derivative of y = x 3 − 48 / x 2 . To find the transition point, interval of increase and decrease, concavity, and asymptotic behavior, we will take the first and second derivative of the given equation as shown below: y = x 3 − 48 x 2 [The given derivative of the equation] y ' = d d x [ x 3 − 48 x 2 ] [Differentiate with respect to x ] y ' = d d x [ x 3 ] − d d x [ 48 x 2 ] [Apply power rule [ u ( x ) n ] ' = n ⋅ u ( x ) n − 1 ] y ' = 3 x 2 − 48 ( − 2 ) x − 3 [ ∴ , d d x [ x 3 ] = 3 x , d d x [ 48 x 2 ] = − 96 x − 3 ] y ' = 3 x 2 + 96 x 3 Find the second derivative by using first derivative are as shown below: y ' = 3 x 2 + 96 x 3 [Find the second derivative] y ' ' = d d x [ 3 x 2 + 96 x 3 ] [Differentiate by the sum rule [ a ⋅ u ( x ) + b ⋅ v ( x ) ] ' = a ⋅ u ' ( x ) + b ⋅ v ' ( x ) ] y ' ' = 3 ⋅ d d x [ x 2 ] + 96 ⋅ d d x [ 1 x 3 ] y ' ' = 3 ⋅ 2 x + 96 ( − 3 ) x − 4 y ' ' = 6 x − 288 x 4 For finding the transition point, the first derivative equal to zero and simplify the critical points

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ISBN: 9781319434526

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 4.5, Problem 35E

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Chapter 4.5, Problem 34E

Chapter 4.5, Problem 36E

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Chapter 4 Solutions

CALCULUS 4E (LL) W/ ACHIEVE ACCESS

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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Chapter 4 Solutions