Explanation Given: The function is f ( x ) = 1 − | x | . Concept used: 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: The given function f ( x ) is an absolute value function. An absolute value function is expressed as shown below: f ( x ) = { 1 − x if x ≥ 0 1 + x if x < 0 For x ≥ 0 the function f ( x ) is shown below: f ( x ) = 1 − x …... (2) Differentiate the function f ( x ) for x ≥ 0 as follows: d d x f ( x ) = d d x ( 1 − x ) f ′ ( x ) = − 1 Substitute c for x in the above equation. f ′ ( c ) = − 1 Substitute b for x in the equation (2) as shown below: f ( b ) = 1 − b Substitute a for x in the equation (2) as shown below: f ( a ) = 1 − a For x ≥ 0 , the value of a and b is greater than 0 or equal to 0 . Now, apply mean value theorem on the interval [ a , b ] for the function f ( x ) = 1 − x . − 1 = ? ( 1 − b ) − ( 1 − a ) b − a − 1 = ? 1 − b − 1 + a b − a − 1 = ? − ( b − a ) b − a − 1 = − 1 From the above calculation it is concluded that if a and b both are positive then the function f ( x ) = 1 − | x | satisfies the MVT

Loose-Leaf Version for Calculus: Early Transcendentals Single Variable

4th Edition

ISBN: 9781319312886

Author: Rogawski, Jon, Adams, Colin, FRANZOSA, Robert

Publisher: W. H. Freeman

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Sequence and Series

A sequence is defined as a collection of things. Series is defined to sum the things one by one in the sequence. It was invented by German mathematician Carl Friedrich Gauss.

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Question

Chapter 4.3, Problem 66E

To determine

To show:

The function f(x)=1−|x| satisfies the mean value theorem on [a,b] if both a and b are negative or positive, but not if a is negative and b is positive.

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Chapter 4.3, Problem 65E

Chapter 4.3, Problem 67E

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Chapter 4 Solutions

Loose-Leaf Version for Calculus: Early Transcendentals Single Variable

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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The function f ( x ) = 1 − | x | satisfies the mean value theorem on [ a , b ] if both a and b are negative or positive, but not if a is negative and b is positive.

Solution Summary: The author explains that the function f(x)=1- left|x-right| satisfies the mean value theorem.

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The function f(x)=1−|x| satisfies the mean value theorem on [a,b] if both a and b are negative or positive, but not if a is negative and b is positive.

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Chapter 4 Solutions

The function f ( x ) = 1 − | x | satisfies the mean value theorem on [ a , b ] if both a and b are negative or positive, but not if a is negative and b is positive.

Solution Summary: The author explains that the function f(x)=1- left|x-right| satisfies the mean value theorem.

Concept explainers

Videos

To show: The function f(x)=1−|x| satisfies the mean value theorem on [a,b] if both a and b are negative or positive, but not if a is negative and b is positive.

Want to see the full answer?

Chapter 4 Solutions

To show:

The function f(x)=1−|x| satisfies the mean value theorem on [a,b] if both a and b are negative or positive, but not if a is negative and b is positive.