Answer
(a) $ f^{-1}(x)=-\frac{1}{x+3}$.
(b) $f$: $(-\infty,0)U(0,\infty)$ and $(-\infty,-3)U(-3,\infty)$.
$f^{-1}$: $(-\infty,-3)U(-3,\infty)$ and $(-\infty,0)U(0,\infty)$.
Work Step by Step
(a) $f(x)=-\frac{3x+1}{x} \longrightarrow y=-\frac{3x+1}{x} \longrightarrow x=-\frac{3y+1}{y}\longrightarrow y=-\frac{1}{x+3}\longrightarrow f^{-1}(x)=-\frac{1}{x+3}$. Check $f(f^{-1})=-\frac{3(-\frac{1}{x+3})+1}{-\frac{1}{x+3}}=x$ and $f^{-1}(f)=-\frac{1}{-\frac{3x+1}{x}+3}=x$
(b) Domain and range of $f$: $(-\infty,0)U(0,\infty)$ and $(-\infty,-3)U(-3,\infty)$. Domain and range of $f^{-1}$: $(-\infty,-3)U(-3,\infty)$ and $(-\infty,0)U(0,\infty)$.
