When does $f^{-1}=\frac{1}{f}$ with $f$ a function mapping $\mathbb{R}^{*}$ to $\mathbb{R}$? In mathematics, an inverse function is a function that "reverses" another function: if the function $f$ applied to an input $x$ gives a result of $y$, then applying its inverse function $g$ to $y$ gives the result $x$, and vice versa. i.e., $f(x) = y$ if and only if $g(y) = x$ , i'm interesting to check the relationship between multiplicative inverse  function and the inverse compositional function ,Then my question here is related to the solution of the below functional equation 


Question:
     When does  $ \displaystyle f^{-1}=\frac{1}{f}$  with $f$ a function mapping $\mathbb{R}^{*}$ to $\mathbb{R}$ and $f(x)\neq 0$ ?
Note: $f^{-1}$ is the compostional inverse of $f$ and $\displaystyle\frac{1}{f}$ is the multiplicative inverse of $f$
Edit: I have edited the question to define $f$ according to the answer below by Robert Israel.


 A: You may like to look into the following article.
Robert Anschuetz II and H. Sherwood, When Is a Function's Inverse Equal to Its Reciprocal? The College Mathematics Journal
Vol. 27, No. 5 (Nov., 1996), pp. 388-393. 
A: You can download a copy of my University of Central Florida thesis that covers this topic in even greater detail than the Journal article below:
http://stars.library.ucf.edu/rtd/3139/
Robert Anschuetz
A: You can't have  $f^{-1}(x) = 1/f(x)$ with both sides defined for all $x \in \mathbb R$.  Namely, if $t = f^{-1}(0)$ then $f^{-1}(t) = 1/f(t) = 1/0$ is undefined.
EDIT:
With the correction that $f$ maps $\mathbb R^* = \mathbb R \backslash \{0\}$ to itself, here is one class of solutions.  Take any $f_0$ that maps $(0,1]$ one-to-one onto $(-\infty,-1]$ with $f_0(1) = -1$.  Then let
$$ f(x) = \cases{f_0(x) & if $x \in (0,1]$\cr
                 1/f_0(1/x) & if $x \in (1,\infty)$\cr
                f_0^{-1}(1/x) & if $x \in (-1,0)$\cr
                 1/f_0^{-1}(x) & if $x \in (-\infty,-1]$\cr} $$
