Chapter 5 - Exponential and Logarithmic Functions - 5.2 One-to-One Functions; Inverse Functions - 5.2 Assess Your Understanding - Page 267: 41

$f(g(x))=f(\frac{1}{x})=\dfrac{1}{\frac{1}{x}}=x$ $g(f(x))=g(\frac{1}{x})=\dfrac{1}{\frac{1}{x}}=x$ $x=1$ has to be excluded because the denominator cannot be $0$ because then the function is undefined.

$f(g(x))=f(\frac{1}{x})=\dfrac{1}{\frac{1}{x}}=x$ $g(f(x))=g(\frac{1}{x})=\dfrac{1}{\frac{1}{x}}=x$ $x=1$ has to be excluded because the denominator cannot be $0$ because then the function is undefined.