Calculus: Graphical, Numerical, Algebraic
Calculus: Graphical, Numerical, Algebraic
3rd Edition
ISBN: 9780133688399
Author: Finney, Ross L.
Publisher: Pearson/Prentice Hall
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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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In calculus, concavity is a descriptor of mathematics that tells about the shape of the graph. It is the parameter that helps to estimate the maximum and minimum value of any of the functions and the concave nature using the graphical method. We use the …
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Chapter 5.3, Problem 52E

a.

To determine

To find: the absolute extrema of f and where they occur.

a.

Expert Solution
Check Mark

Answer to Problem 52E

The absolute maximum is (0,2) and absolute minimum is (−2,−1)(2,−1)

Explanation of Solution

Given information: f is an even function, continuous [−3,3] and satisfies the following:

    x012
    f20−1
    f’Does not exist0Does not exist
    f’’Does not exist0Does not exist
    x0 1 2
    f+−−
    f’−−+
    f’’+−−

f is an even function; therefore, the table can be extended for the values of x between −3 and 0. Since it is even the portion on the left will be the mirror image of the portion on the right, that is, the y -coordinates remain same, the slopes have opposite sign and the concavity remains same.

    x−2−1012
    f−1020−1
    f’Does not exist0Does not exist0Does not exist
    f’’Does not exist0Does not exist0Does not exist
    x−3 −2 −1 0 1 2
    f−−++−−
    f’−++−−+
    f’’−−++−−

  f' changes from positive to negative at x = 0, therefore it is a maximum.

Since, f < 0 for −3 < x < −2 and 2 x=3 and x=−3 is negative and is less than 2, x=2 , therefore, (0,2) is the absolute maximum.

  f' changes from negative to positive at x=−2 and x=2 , therefore, they are minimums.

Since, f < 0 for −3 f(−3)>f(−2) and f > 0 for 2 f(3)>f(2) , therefore, they are the absoluteminimums

Hence, the absolute maximum is (0,2) and absolute minimum is (−2,−1)(2,−1)

b.

To determine

To find: the points of inflection.

b.

Expert Solution
Check Mark

Answer to Problem 52E

The points of inflection are x=±1 .

Explanation of Solution

Given information: f is an even function, continuous [−3,3] and satisfies the following:

    X012
    F20−1
    f’Does not exist0Does not exist
    f’’Does not exist0Does not exist
    X0 1 2
    F+−−
    f’−−+
    f’’+−−

  f'' shows the concavity of the function which determines the points of inflection or the points where the concavity of the function changes.

  f''=0 at f = 1 the concavity changes and also at f = −1 since, it is an even function.

Hence, the points of inflection are x=±1 .

c.

To determine

To graph: the function f with the provided information.

c.

Expert Solution
Check Mark

Explanation of Solution

Given information: f is an even function, continuous [−3,3] and satisfies the following:

    X012
    F20−1
    f’Does not exist0Does not exist
    f’’Does not exist0Does not exist
    X0 1 2
    F+−−
    f’−−+
    f’’+−−

Graph:

  Calculus: Graphical, Numerical, Algebraic, Chapter 5.3, Problem 52E

Interpretation:

Hence, this is the probable graph for function f with the provided information.

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Chapter 5.3, Problem 51E
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Chapter 5.3, Problem 53E

Chapter 5 Solutions

Calculus: Graphical, Numerical, Algebraic

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