Explanation Given: The function is f ( x ) = m x + b . The double differentiation f ″ exists and f ″ ( x ) = 0 for all x . Calculation: The given equation is as follows: f ″ ( x ) = 0 ….. (1) Integrate equation (1) as shown below: ∫ f ″ ( x ) d x = ∫ 0 d x + m f ′ ( x ) = m Label the above equation as equation (2). f ′ ( x ) = m …. (2) In the above equation m is a constant

Loose-leaf Version For Calculus: Early Transcendentals

4th Edition

ISBN: 9781319055912

Author: Jon Rogawski, Colin Adams, Robert Franzosa

Publisher: W. H. Freeman

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Differential Equations

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Chapter 4.3, Problem 77E

To determine

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The function f(x) is expressed as f(x)=mx+b, where m=f′(0) and b=f(0) such that f″(x)=0 for all x.

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Chapter 4.3, Problem 76E

Chapter 4.3, Problem 78E

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Chapter 4 Solutions

Loose-leaf Version For Calculus: Early Transcendentals

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The function f ( x ) is expressed as f ( x ) = m x + b , where m = f ′ ( 0 ) and b = f ( 0 ) such that f ″ ( x ) = 0 for all x .

Solution Summary: The author explains how the function f(x) is expressed in mx+b.

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The function f(x) is expressed as f(x)=mx+b, where m=f′(0) and b=f(0) such that f″(x)=0 for all x.

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Chapter 4 Solutions

The function f ( x ) is expressed as f ( x ) = m x + b , where m = f ′ ( 0 ) and b = f ( 0 ) such that f ″ ( x ) = 0 for all x .

Solution Summary: The author explains how the function f(x) is expressed in mx+b.

Concept explainers

Videos

To show: The function f(x) is expressed as f(x)=mx+b, where m=f′(0) and b=f(0) such that f″(x)=0 for all x.

Want to see the full answer?

Chapter 4 Solutions

To show:

The function f(x) is expressed as f(x)=mx+b, where m=f′(0) and b=f(0) such that f″(x)=0 for all x.