a. Answer The absolute maximum is ( 1 , 2 ) and absolute minimum is ( 3 , − 2 ) Explanation Given information : f is continuous on [ 0 , 3 ] and satisfies the following: x 0 1 2 3 f 1 2 0 −2 f’ 3 0 Does not exist −3 f’’ 0 −1 Does not exist 0 x 0 < x < 1 1 < x < 2 2 < x < 3 f + + − f’ + − − f’’ − − − The tables shows that, f ( 1 ) = 2 f ( 3 ) = − 2 The derivative of 1 is 0, f ' changes from positive to negative at x = 1, therefore it is a maximum because f ' is negative for the rest of the domain, ( 1 , 3 ] . ( 3 , − 2 ) is a minimum as there is no other value that is below this point. Hence, the absolute maximum is ( 1 , 2 ) and absolute minimum is ( 3 , − 2 ) b. To determine To find : the points of inflection. Answer There are no points of inflection. Explanation Given information : f is continuous on [ 0 , 3 ] and satisfies the following: x 0 1 2 3 f 1 2 0 −2 f’ 3 0 Does not exist −3 f’’ 0 −1 Does not exist 0 x 0 < x < 1 1 < x < 2 2 < x < 3 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. The provided information shows that the function is always concave down, that is f ' ' is always negative. Hence, there are no points of inflection. c. To determine To graph : the function f with the provided information. Explanation Given information : f is continuous on [ 0 , 3 ] and satisfies the following: x 0 1 2 3 f 1 2 0 −2 f’ 3 0 Does not exist −3 f’’ 0 −1 Does not exist 0 X 0 < x < 1 1 < x < 2 2 < x < 3 F + + − f’ + − − f’’ − − − Graph : Interpretation: Hence, this is the probable graph for function f with the provided information.

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

4th Edition

ISBN: 9780133178579

Author: Ross L. Finney

Publisher: PEARSON

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Question

Chapter 5.3, Problem 51E

To determine

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

Expert Solution

Answer to Problem 51E

The absolute maximum is (1,2) and absolute minimum is (3,−2)

Explanation of Solution

Given information: f is continuous on [0,3] and satisfies the following:

x	0	1	2	3
f	1	2	0	−2
f’	3	0	Does not exist	−3
f’’	0	−1	Does not exist	0

x	0 < x < 1	1 < x < 2	2 < x < 3
f	+	+	−
f’	+	−	−
f’’	−	−	−

The tables shows that,

f(1)=2f(3)=−2

The derivative of 1 is 0, f' changes from positive to negative at x = 1, therefore it is a maximum because f' is negative for the rest of the domain, (1,3] .

(3,−2) is a minimum as there is no other value that is below this point.

Hence, the absolute maximum is (1,2) and absolute minimum is (3,−2)

To determine

To find: the points of inflection.

Expert Solution

Answer to Problem 51E

There are no points of inflection.

Explanation of Solution

Given information: f is continuous on [0,3] and satisfies the following:

x	0	1	2	3
f	1	2	0	−2
f’	3	0	Does not exist	−3
f’’	0	−1	Does not exist	0

x	0 < x < 1	1 < x < 2	2 < x < 3
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.

The provided information shows that the function is always concave down, that is f'' is always negative.

Hence, there are no points of inflection.

To determine

To graph: the function f with the provided information.

Expert Solution

Explanation of Solution

Given information: f is continuous on [0,3] and satisfies the following:

x	0	1	2	3
f	1	2	0	−2
f’	3	0	Does not exist	−3
f’’	0	−1	Does not exist	0

X	0 < x < 1	1 < x < 2	2 < x < 3
F	+	+	−
f’	+	−	−
f’’	−	−	−

Graph:

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 5.3, Problem 51E

Interpretation:

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

Chapter 5.3, Problem 50E

Chapter 5.3, Problem 52E

Chapter 5 Solutions

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

Ch. 5.1 - Prob. 1QR Ch. 5.1 - Prob. 2QR Ch. 5.1 - Prob. 3QR Ch. 5.1 - Prob. 4QR Ch. 5.1 - Prob. 5QR Ch. 5.1 - Prob. 6QR Ch. 5.1 - Prob. 7QR Ch. 5.1 - Prob. 8QR Ch. 5.1 - Prob. 9QR Ch. 5.1 - Prob. 10QR

Ch. 5.1 - Prob. 11QR Ch. 5.1 - Prob. 12QR Ch. 5.1 - Prob. 1E Ch. 5.1 - Prob. 2E Ch. 5.1 - Prob. 3E Ch. 5.1 - Prob. 4E Ch. 5.1 - Prob. 5E Ch. 5.1 - Prob. 6E Ch. 5.1 - Prob. 7E Ch. 5.1 - Prob. 8E Ch. 5.1 - Prob. 9E Ch. 5.1 - Prob. 10E Ch. 5.1 - Prob. 11E Ch. 5.1 - Prob. 12E Ch. 5.1 - Prob. 13E Ch. 5.1 - Prob. 14E Ch. 5.1 - Prob. 15E Ch. 5.1 - Prob. 16E Ch. 5.1 - Prob. 17E Ch. 5.1 - Prob. 18E Ch. 5.1 - Prob. 19E Ch. 5.1 - Prob. 20E Ch. 5.1 - Prob. 21E Ch. 5.1 - Prob. 22E Ch. 5.1 - Prob. 23E Ch. 5.1 - Prob. 24E Ch. 5.1 - Prob. 25E Ch. 5.1 - Prob. 26E Ch. 5.1 - Prob. 27E Ch. 5.1 - Prob. 28E Ch. 5.1 - Prob. 29E Ch. 5.1 - Prob. 30E Ch. 5.1 - Prob. 31E Ch. 5.1 - Prob. 32E Ch. 5.1 - Prob. 33E Ch. 5.1 - Prob. 34E Ch. 5.1 - Prob. 35E Ch. 5.1 - Prob. 36E Ch. 5.1 - Prob. 37E Ch. 5.1 - Prob. 38E Ch. 5.1 - Prob. 39E Ch. 5.1 - Prob. 40E Ch. 5.1 - Prob. 41E Ch. 5.1 - Prob. 42E Ch. 5.1 - Prob. 43E Ch. 5.1 - Prob. 44E Ch. 5.1 - Prob. 45E Ch. 5.1 - Prob. 46E Ch. 5.1 - Prob. 47E Ch. 5.1 - Prob. 48E Ch. 5.1 - 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