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Found in: Page 216

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# 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.

See the step by step solution

## 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.