Chapter 3 - Applications of Differentiation - 3.2 The Mean Value Theorem - 3.2 Exercises - Page 220: 33

No

For $x$ $\gt$ $0$, $f(x)$ = $g(x)$ so $f'(x)$ = $g'(x)$. For $x$ $\lt$ $0$, $f'(x)$ = $(\frac{1}{x})'$ = $-\frac{1}{x^{2}}$ and $g'(x)$ = $(1+\frac{1}{x})'$ = $-\frac{1}{x^{2}}$ so $f'(x)$ = $g'(x)$ However the domain of $f$ and $g$ is not an interval, it is $(-\infty,0)$ $∪$ $(0,\infty)$ so we cannot conclude that $f-g$ is constant.