Explanation Given information:- The given derivative of y = x 3 − 4 x 2 + 1 . Calculations:- The given derivative of y = x 3 − 4 x 2 + 1 . To find the transition point, we will take the first and second derivative of the given equation as shown below: y = x 3 − 4 x 2 + 1 [The given derivative of the equation] y ' = d d x [ x 3 − 4 x 2 + 1 ] [Differentiate with respect to x ] y ' = − 4 ⋅ d d x [ 1 x 2 + 1 ] + d d x [ x 3 ] [Apply power rule [ u ( x ) n ] ' = n ⋅ u ( x ) n − 1 ] y ' = 3 x 2 +4 ⋅ d d x [ x 2 + 1 ] ( x 2 + 1 ) 2 [ ∴ , d d x [ x 3 ] = 3 x 2 ] y ' = 4 ( d d x [ x 2 ] + d d x [ 1 ] ) ( x 2 + 1 ) 2 + 3 x 2 [Apply the quotient rule [ u ( x ) v ( x ) ] ' = u ' ( x ) ⋅ v ( x ) − u ( x ) ⋅ v ' ( x ) v ( x ) 2 ] y ' = 4 ( 2 x + 0 ) ( x 2 + 1 ) 2 + 3 x 2 y ' = 8 x ( x 2 + 1 ) 2 + 3 x 2 Find the second derivative by using first derivative are as shown below: y ' = 8 x ( x 2 + 1 ) 2 + 3 x 2 [Find the second derivative] y ' ' = d d x [ 8 x ( x 2 + 1 ) 2 + 3 x 2 ] [Apply the quotient rule [ u ( x ) v ( x ) ] ' = u ' ( x ) ⋅ v ( x ) − u ( x ) ⋅ v ' ( x ) v ( x ) 2 ] y ' ' = 8 ⋅ d d x [ x ( x 2 + 1 ) 2 ] + 3 ⋅ d d x [ x 2 ] y ' ' = 8 ⋅ d d x [ x ] ⋅ ( x 2 + 1 ) 2 − x ⋅ d d x [ ( x 2 + 1 ) 2 ] ( ( x 2 + 1 ) 2 ) 2 + 3 ⋅ 2 x y ' ' = 8 ⋅ ( 1 ( x 2 + 1 ) 2 − 2 ( x 2 + 1 ) ⋅ d d x [ x 2 + 1 ] ⋅ x ) ( x 2 + 1 ) 4 + 6 x [Simplify] y ' ' = 8 ⋅ ( ( x 2 + 1 ) 2 − 2 ( d d x [ x 2 ] + d d x [

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Domain

In simple mathematics terms, a domain is defined as the space on which all the possible inputs for the function will lie. It can also be expressed as a set of departures for the required function.

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Chapter 4.5, Problem 39E

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The solution of the graph of the equation and indicate all the transition points.

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Chapter 4.5, Problem 38E

Chapter 4.5, Problem 40E

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Chapter 4 Solutions

Calculus - Standalone book

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

Ch. 4.1 - Prob. 7E Ch. 4.1 - Prob. 8E Ch. 4.1 - Prob. 9E Ch. 4.1 - Prob. 10E Ch. 4.1 - Prob. 11E Ch. 4.1 - Prob. 12E Ch. 4.1 - Prob. 13E Ch. 4.1 - Prob. 14E Ch. 4.1 - Prob. 15E Ch. 4.1 - Prob. 16E Ch. 4.1 - Prob. 17E Ch. 4.1 - Prob. 18E Ch. 4.1 - Prob. 19E Ch. 4.1 - Prob. 20E Ch. 4.1 - Prob. 21E Ch. 4.1 - Prob. 22E Ch. 4.1 - Prob. 23E Ch. 4.1 - Prob. 24E Ch. 4.1 - Prob. 25E Ch. 4.1 - Prob. 26E Ch. 4.1 - Prob. 27E Ch. 4.1 - Prob. 28E Ch. 4.1 - Prob. 29E Ch. 4.1 - Prob. 30E Ch. 4.1 - Prob. 31E Ch. 4.1 - Prob. 32E Ch. 4.1 - Prob. 33E Ch. 4.1 - Prob. 34E Ch. 4.1 - Prob. 35E Ch. 4.1 - Prob. 36E Ch. 4.1 - Prob. 37E Ch. 4.1 - Prob. 38E Ch. 4.1 - Prob. 39E Ch. 4.1 - Prob. 40E Ch. 4.1 - Prob. 41E Ch. 4.1 - Prob. 42E Ch. 4.1 - Prob. 43E Ch. 4.1 - Prob. 44E Ch. 4.1 - Prob. 45E Ch. 4.1 - Prob. 46E Ch. 4.1 - Prob. 47E Ch. 4.1 - Prob. 48E Ch. 4.1 - Prob. 49E Ch. 4.1 - Prob. 50E Ch. 4.1 - Prob. 51E Ch. 4.1 - Prob. 52E Ch. 4.1 - Prob. 53E Ch. 4.1 - Prob. 54E Ch. 4.1 - Prob. 55E Ch. 4.1 - Prob. 56E Ch. 4.1 - 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The solution of the graph of the equation and indicate all the transition points.

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Chapter 4 Solutions