Question

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?

Short answer

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

Step-by-step solution

Given data

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

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".

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.

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.