# 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.

• Your edit fails to dispense with Robert Israel's counterexample. – Steven Landsburg Jan 26 '17 at 1:51
• – Watson Jan 26 '17 at 11:30
• – Watson Jan 26 '17 at 12:37
• [I have edited the question .../... Robert Israel.]---> then you should better put $f:\mathbb{R}^*\to \mathbb{R}^*$ – Duchamp Gérard H. E. Mar 4 '17 at 3:32

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 $$f(x) = \cases{f_0(x) & if x \in (0,1]\cr 1/f_0(1/x) & if x \in (1,\infty)\cr f_0^{-1}(1/x) & if x \in (-1,0)\cr 1/f_0^{-1}(x) & if x \in (-\infty,-1]\cr}$$

• Thanks for that , just i forgot that , i'm going to edit it – zeraoulia rafik Jan 26 '17 at 0:25
• For example, $f_0(x)=-1+\tan\left(\frac{(x-1)\pi}2\right)$. – T. Amdeberhan Jan 26 '17 at 2:46

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.

• Thanks for the paper but unfourtinaltely it's not free – zeraoulia rafik Jan 26 '17 at 0:22
• (1) You may ask the authors; or, (2) Someone here might have access copy; or. (3) Will try to get you tomorrow. – T. Amdeberhan Jan 26 '17 at 0:26
• @zeraouliarafik If only there were some sort of scientific hub where many behind-a-paywall papers were available. (It has this paper!) – dvitek Jan 26 '17 at 0:53
• Thanks for this orientation , really i have checked bookzz.org but i don't find it available – zeraoulia rafik Jan 26 '17 at 0:55
• @MattF. only if you sign up, which means it's not really freely available. But I get your point. – David Roberts Jan 26 '17 at 5:36

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

• Thanks , your paper is already montioned and cited above by the answer of T. Amdeberhan – zeraoulia rafik Mar 3 '17 at 21:16