Explanation Given information:- The given derivative of y = x 2 − x + x − 1 . Calculations:- The given derivative of y = x 2 − x + x − 1 . 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 2 − x + x − 1 [The given derivative of the equation] y ' = d d x [ x 2 − x + x − 1 . ] [Differentiate with respect to x ] y ' = d d x [ x 2 ] − d d x [ x ]+ d d x [ 1 x ] [Apply power rule [ u ( x ) n ] ' = n ⋅ u ( x ) n − 1 ] y ' = 2 x − 1 − d d x [ x ] x 2 [ ∴ , d d x [ x 2 ] = 2 x , d d x [ x ] = 1 ] y ' = 2 x − 1 − 1 x 2 Find the second derivative by using first derivative are as shown below: y ' = 2 x − 1 − 1 x 2 [Find the second derivative] y ' ' = d d x [ 2 x − 1 − 1 x 2 ] [Differentiate by the sum rule [ a ⋅ u ( x ) + b ⋅ v ( x ) ] ' = a ⋅ u ' ( x ) + b ⋅ v ' ( x ) ] y ' ' = 2 ⋅ d d x [ x ] − d d x [ 1 x 2 ] + d d x [ − 1 ] y ' ' = 2 x 3 + 2 For finding the transition point, the first derivative equal to zero and simplify the critical points

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ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 4.5, Problem 36E

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Chapter 4.5, Problem 35E

Chapter 4.5, Problem 37E

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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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