## Precalculus (10th Edition)

$f(g(x))=f(\frac{1}{x})=\dfrac{1}{\frac{1}{x}}=x$ $g(f(x))=g(\frac{1}{x})=\dfrac{1}{\frac{1}{x}}=x$ $x=1$ 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.