# Let f:Rise defined from R to R . find the solution of f(x) =f^-1 (x)?

May 27, 2018

$f \left(x\right) = x$

#### Explanation:

We seek a function $f : \mathbb{R} \rightarrow \mathbb{R}$ such that solution $f \left(x\right) = {f}^{- 1} \left(x\right)$

That is we seek a function that is its own inverse. One obvious such function is the trivial solution:

$f \left(x\right) = x$

However, a more thorough analysis of the problem is of significant complexity as explored by Ng Wee Leng and Ho Foo Him as published in the Journal of the Association of Teachers of Mathematics.

https://www.atm.org.uk/journal/archive/mt228files/atm-mt228-39-42.pdf

May 27, 2018

Check below.

#### Explanation:

The points in common between ${C}_{f}$ and ${C}_{{f}^{- 1}}$ if they exist they are not always in the bisector $y = x$. Here is an example of such a function: $f \left(x\right) = 1 - {x}^{2}$ $\textcolor{w h i t e}{a}$ , $x$$\in$$\left[0 , + \infty\right)$

graph{((y-(1-x^2))sqrtx)=0 [-7.02, 7.03, -5.026, 1.994]}

They are however in the bisector only and only if $f$ is ↗ increasing.

If $f$ is strictly increasing then $f \left(x\right) = {f}^{- 1} \left(x\right)$ $\iff$ $f \left(x\right) = x$

If $f$ is not strictly increasing the common points are found by solving the system of equations

$\left\{\begin{matrix}y = f \left(x\right) \text{ " \\ x=f^(-1)(y)" }\end{matrix}\right.$ $\iff$ $\left\{\begin{matrix}y = f \left(x\right) \text{ " \\ x=f(y)" }\end{matrix}\right.$ $\iff \ldots$

May 28, 2018

${f}^{- 1} \left(x\right) = f \left(x\right)$ $\iff x = 1$

#### Explanation:

$f \left(x\right) = {x}^{3} + x - 1$ $\textcolor{w h i t e}{a a}$, $x$$\in$$\mathbb{R}$

$f ' \left(x\right) = 3 {x}^{2} + 1 > 0$ $\textcolor{w h i t e}{a a}$ , $\forall$$x$$\in$$\mathbb{R}$

so $f$ is ↗ in $\mathbb{R}$. As a strictly monotone function it is also "$1 - 1$" and as a one to one function it has an inverse.

We need to solve the equation ${f}^{- 1} \left(x\right) = f \left(x\right)$ <=>^(f↗)f(x)=x $\iff$

${x}^{3} + x - 1 = x$ $\iff$ ${x}^{3} - 1 = 0$ $\iff$

$\left(x - 1\right) \left({x}^{2} + x + 1\right) = 0$ ${\iff}^{{x}^{2} + x + 1 > 0}$

$x = 1$