Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

Let \(f(x) = \frac{1}{x}\) and

**\(g(x) = \left\{ {\begin{aligned}{*{20}{l}}{\frac{1}{x}}&{ if x > 0}\\{1 + \frac{1}{x}}&{ if x > 0}\end{aligned}} \right.\)

Show that \({f^\prime }(x) = {g^\prime }(x)\) for all \(x\) in their domains. Can we conclude from Corollary 7 that \(f - g\) is constant?**

The \(f - g\) is not constant over the whole domain. It contradicts corollary that \(f - g\) is constant although the derivatives are equal.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

The functions \(f(x) = \frac{1}{x}\) and \(g(x) = \left\{ {\begin{aligned}{*{20}{l}}{\frac{1}{x}}&{{\rm{ if }}x > 0}\\{1 + \frac{1}{x}}&{{\rm{ if }}x > 0}\end{aligned}} \right.\)

Step 2: Concept of corollary

"If \({f^\prime }(x) = {g^\prime }(x)\) for all \(x\) in an interval \((a,\;b)\), then \(f - g\) is constant on \((a,\;b)\), that is \(f(x) = g(x) + c\) where \(c\) is constant".

Step 3: Show the statement \({f^\prime }(x) = {g^\prime }(x)\) for all \(x\) in their domains

Obtain the derivative of \(f(x)\) as follows:

\(\begin{aligned}{c}{f^\prime }(x) &= \frac{d}{{dx}}\left( {\frac{1}{x}} \right)\\ &= \frac{{ - 1}}{{{x^2}}}\end{aligned}\)

Obtain the derivative of \(g(x)\) for \(x > 0\) as follows:

\(\begin{aligned}{c}{g^\prime }(x) &= \frac{d}{{dx}}\left( {\frac{1}{x}} \right)\\ &= \frac{{ - 1}}{{{x^2}}}\end{aligned}\)

Obtain the derivative of \(g(x)\) for \(x < 0\) as follows:

\(\begin{aligned}{c}{g^\prime }(x) &= \frac{d}{{dx}}\left( {1 + \frac{1}{x}} \right)\\ &= 0 + \left( {\frac{{ - 1}}{{{x^2}}}} \right)\\ &= \frac{{ - 1}}{{{x^2}}}\end{aligned}\)

For the case \(x > 0\), from the above equations, \({f^\prime }(x) = {g^\prime }(x)\) and for the case \(x < 0\), from the above equations, \({f^\prime }(x) = {g^\prime }(x)\).

Hence, the statement is showed for all \(x\) in their domains.

Step 4: Justify \(f - g\) is a constant

Subtract the given functions as follows:

For the case of \(x > 0,f - g\) will be as follows:

\(\begin{aligned}{c}f(x) - g(x) &= \frac{1}{x} - \frac{1}{x}\\ &= 0\end{aligned}\)

For the case of \(x < 0,f - g\) will be as follows:

\(\begin{aligned}{c}f(x) - g(x) &= \frac{1}{x} - \left( {1 + \frac{1}{x}} \right)\\ &= \frac{1}{x} - 1 - \frac{1}{x}\\ &= - 1\end{aligned}\)

Hence, it can be concluded that \(f - g\) is not constant over the whole domain. It contradicts corollary that \(f - g\) is constant although the derivatives are equal.

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Let \(f(x) = \frac{1}{x}\) and

**\(g(x) = \left\{ {\begin{aligned}{*{20}{l}}{\frac{1}{x}}&{ if x > 0}\\{1 + \frac{1}{x}}&{ if x > 0}\end{aligned}} \right.\)

Show that \({f^\prime }(x) = {g^\prime }(x)\) for all \(x\) in their domains. Can we conclude from Corollary 7 that \(f - g\) is constant?**

Step by Step Solution

Step 1: Given data

Step 2: Concept of corollary

Step 3: Show the statement \({f^\prime }(x) = {g^\prime }(x)\) for all \(x\) in their domains

Step 4: Justify \(f - g\) is a constant

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Let \(f(x) = \frac{1}{x}\) and \(g(x) = \left\{ {\begin{aligned}{*{20}{l}}{\frac{1}{x}}&{ if x > 0}\\{1 + \frac{1}{x}}&{ if x > 0}\end{aligned}} \right.\) Show that \({f^\prime }(x) = {g^\prime }(x)\) for all \(x\) in their domains. Can we conclude from Corollary 7 that \(f - g\) is constant?

Step by Step Solution

Step 1: Given data

Step 2: Concept of corollary

Step 3: Show the statement \({f^\prime }(x) = {g^\prime }(x)\) for all \(x\) in their domains

Step 4: Justify \(f - g\) is a constant

Most popular questions for Math Textbooks

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Pure Maths

Let \(f(x) = \frac{1}{x}\) and

**\(g(x) = \left\{ {\begin{aligned}{*{20}{l}}{\frac{1}{x}}&{ if x > 0}\\{1 + \frac{1}{x}}&{ if x > 0}\end{aligned}} \right.\)

Show that \({f^\prime }(x) = {g^\prime }(x)\) for all \(x\) in their domains. Can we conclude from Corollary 7 that \(f - g\) is constant?**