Explanation Given: f ( x ) = tan − 1 x 2 . Calculation: Finding the root of the function: f ( x ) = 0 t a n − 1 x 2 = 0 x 2 = tan 0 x = 0 The first derivative of the function is given below: f ' ( x ) = d d x ( tan − 1 x 2 ) f ' ( x ) = 1 1 + x 2 .2 x f ' ( x ) = 2 x 1 + x 2 This derivatives is always defined and continuous, so the critical points of are just the places where f ' ( x ) = 0 ; that is 2 x 1 + x 2 = 0 2 x = 0 x = 0 Therefore, the function has a critical point at x = 0 . Now, f ( 0 ) = tan − 1 0 2 f ( 0 ) = tan − 1 0 2 f ( 0 ) = tan − 1 0 f ( 0 ) = 0 Therefore, the function f ( x ) has a local minimum at x = 0

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Author: Laura Taalman, Peter Kohn

Publisher: W. H. Freeman

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Maxima and Minima

The extreme points of a function are the maximum and the minimum points of the function. A maximum is attained when the function takes the maximum value and a minimum is attained when the function takes the minimum value.

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Chapter 3, Problem 16SC

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To find: The roots, local maxima and minima and inflection points and also determine whether any local extrema are also global extrema on the domain for the given function.

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Chapter 3, Problem 15SC

Chapter 3, Problem 17SC

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How would i solve this. More info is that b =1 but it might be better to solve this before making the substitution

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Chapter 3 Solutions

Calculus

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The roots, local maxima and minima and inflection points and also determine whether any local extrema are also global extrema on the domain for the given function.

Solution Summary: The author calculates the root, local maxima and minima, inflection points, and global extrema for the given function.

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To find: The roots, local maxima and minima and inflection points and also determine whether any local extrema are also global extrema on the domain for the given function.

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Chapter 3 Solutions