Explanation Given information:- The given derivative of y = ( x 3 − 3 x ) 1 / 3 . Calculations:- The given derivative of y = ( x 3 − 3 x ) 1 / 3 . 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 − 3 x ) 1 / 3 . [The given derivative of the equation] y ' = d d x ( x 3 − 3 x ) 1 / 3 [Differentiate with respect to x ] y ' = 1 3 ( x 3 − 3 x ) 1 3 − 1 ⋅ d d x ( x 3 − 3 x ) [Apply power rule [ u ( x ) n ] ' = n ⋅ u ( x ) n − 1 ⋅ u ' ( x ) ] y ' = d d x [ x 3 ] − 3 ⋅ d d x [ x ] 3 ( x 3 − 3 x ) 2 3 [ ∴ , d d x [ x ] = 1 , d d x [ x 2 ] = 2 x ] y ' = 3 x 2 − 3 3 ( x 3 − 3 x ) 2 3 = x 2 − 1 ( x 3 − 3 x ) 2 3 Find the second derivative by using first derivative are as shown below: y ' = x 2 − 1 ( x 3 − 3 x ) 2 3 [Find the second derivative] y ' ' = d d x [ x 2 − 1 ( x 3 − 3 x ) 2 3 ] [Differentiate by the quotient rule [ u ( x ) v ( x ) ] ' = u ' ( x ) ⋅ v ( x ) − u ( x ) ⋅ v ' ( x ) v ( x ) 2 ] y ' ' = d d x [ x 2 − 1 ] ⋅ ( x 2 − 3 x ) 2 3 − ( x 2 − 1 ) ⋅ d d x [ ( x 2 − 3 x ) 2 3 ] ( ( x 2 − 3 x ) 2 3 ) 2 y " = ( d d x [ x 2 ] + d d x [ − 1 ] ) ( x 2 − 3 x ) 2 3 − 2 3 ( x 2 − 3 x ) 2 3 − 1 ⋅ d d x [ x 2 − 3 x ] ( x 2 − 3 x ) 4 3 y " =

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ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 4.5, Problem 32E

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Chapter 4.5, Problem 31E

Chapter 4.5, Problem 33E

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Chapter 4 Solutions

CALCULUS W/SAPLING ACCESS >IC<

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