Explanation Given: The value of f ( 2 ) is − 2 and f ′ ( x ) ≥ 5 . Concept used: The mean value theorem: The mean value theorem states that if the function f is continuous on the interval [ a , b ] and differentiable on the interval ( a , b ) then a number c exists in the interval such that the slope at c can be expressed as follows: f ′ ( c ) = f ( b ) − f ( a ) b − a ...... (1) Proof: Mean value theorem is applied on a function which is differentiable on the interval ( a , b ) such that the value of a and b are well defined. So, for the inequality x > 2 the corresponding interval is ( 2 , x ) . Now, the mean value theorem can be applied on the interval [ 2 , x ] . Substitute x for b , 2 for a and x for c in equation (1) as follows: f ′ ( x ) = f ( x ) − f ( 2 ) x − 2 It is given that f ′

4th Edition

ISBN: 9781319050740

Author: Jon Rogawski, Colin Adams, Robert Franzosa

Publisher: W. H. Freeman

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Question

Chapter 4.3, Problem 70E

To determine

To show:

That f(4)≥8 for all x>2 if f(2)=−2 and f′(x)≥5.

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Chapter 4.3, Problem 69E

Chapter 4.3, Problem 71E

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Chapter 4 Solutions

Calculus: Early Transcendentals

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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Prob. 3E Ch. 4.8 - Prob. 4E Ch. 4.8 - Prob. 5E Ch. 4.8 - Prob. 6E Ch. 4.8 - Prob. 7E Ch. 4.8 - Prob. 8E Ch. 4.8 - Prob. 9E Ch. 4.8 - Prob. 10E Ch. 4.8 - Prob. 11E Ch. 4.8 - Prob. 12E Ch. 4.8 - Prob. 13E Ch. 4.8 - Prob. 14E Ch. 4.8 - Prob. 15E Ch. 4.8 - Prob. 16E Ch. 4.8 - Prob. 17E Ch. 4.8 - Prob. 18E Ch. 4.8 - Prob. 19E Ch. 4.8 - Prob. 20E Ch. 4.8 - Prob. 21E Ch. 4.8 - Prob. 22E Ch. 4.8 - Prob. 23E Ch. 4.8 - Prob. 24E Ch. 4.8 - Prob. 25E Ch. 4.8 - Prob. 26E Ch. 4.8 - Prob. 27E Ch. 4.8 - Prob. 28E Ch. 4.8 - Prob. 29E Ch. 4.8 - Prob. 30E Ch. 4.8 - Prob. 31E Ch. 4.8 - Prob. 32E Ch. 4.8 - Prob. 33E Ch. 4.8 - Prob. 34E Ch. 4.8 - Prob. 35E Ch. 4.8 - Prob. 36E Ch. 4.8 - Prob. 37E Ch. 4.8 - Prob. 38E Ch. 4.8 - Prob. 39E Ch. 4.8 - Prob. 40E Ch. 4.8 - Prob. 41E Ch. 4 - Prob. 1CRE Ch. 4 - Prob. 2CRE Ch. 4 - Prob. 3CRE Ch. 4 - Prob. 4CRE Ch. 4 - Prob. 5CRE Ch. 4 - Prob. 6CRE Ch. 4 - Prob. 7CRE Ch. 4 - Prob. 8CRE Ch. 4 - Prob. 9CRE Ch. 4 - Prob. 10CRE Ch. 4 - Prob. 11CRE Ch. 4 - Prob. 12CRE Ch. 4 - 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Prob. 63CRE Ch. 4 - Prob. 64CRE Ch. 4 - Prob. 65CRE Ch. 4 - Prob. 66CRE Ch. 4 - Prob. 67CRE Ch. 4 - Prob. 68CRE Ch. 4 - Prob. 69CRE Ch. 4 - Prob. 70CRE Ch. 4 - Prob. 71CRE Ch. 4 - Prob. 72CRE Ch. 4 - Prob. 73CRE Ch. 4 - Prob. 74CRE Ch. 4 - Prob. 75CRE Ch. 4 - Prob. 76CRE Ch. 4 - Prob. 77CRE Ch. 4 - Prob. 78CRE Ch. 4 - Prob. 79CRE Ch. 4 - Prob. 80CRE Ch. 4 - Prob. 81CRE Ch. 4 - Prob. 82CRE Ch. 4 - Prob. 83CRE Ch. 4 - Prob. 84CRE Ch. 4 - Prob. 85CRE Ch. 4 - Prob. 86CRE Ch. 4 - Prob. 87CRE Ch. 4 - Prob. 88CRE Ch. 4 - Prob. 89CRE

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