We have, $f:[1, \infty) \rightarrow[5, \infty)$ given by $f(x)=3 x+\frac{2}{x}$
Let $\quad y=f(x)=3 x+\frac{2}{x}$.
Then, we get
$$
\begin{aligned}
& 3 x^2+2-y x=0 \Rightarrow 3 x^2-y x+2=0 \\
\Rightarrow & x=\frac{y \pm \sqrt{y^2-4 \times 2 \times 3}}{6} \\
\Rightarrow & x=\frac{y \pm \sqrt{y^2-24}}{6}
\end{aligned}
$$
Now, as $\quad 5 \leq y < \infty$
$$
\begin{array}{rr}
\therefore & 25 \leq y^2 < \infty \Rightarrow 1 \leq y^2-24 < \infty \\
\Rightarrow & 1 \leq \sqrt{y^2-24} < \infty \\
\Rightarrow & 6 \leq y+\sqrt{y^2-24} < \infty \\
\Rightarrow & 1 \leq \frac{1}{6}\left(y+\sqrt{y^2-24}\right) < \infty \\
\Rightarrow & 1 \leq x < \infty \Rightarrow x \in[1, \infty)
\end{array}
$$
[Note that $x=\frac{y-\sqrt{y^2-24}}{6} \notin[1, \infty) \forall y \in[5, \infty]$.


$$
\begin{array}{ll}
\therefore & f^{-1}(y)=\frac{y+\sqrt{y^2-24}}{6} \\
\text { or } & f^{-1}(x)=\frac{x+\sqrt{x^2-24}}{6}
\end{array}
$$