Unseen errors

The no of solutions of the equation $f(x)= f{-1}(x)$ should occur for the ordered pairs of type ( a , a ) algebrically which gives us the equation
$({ x }^{ 5 }-x+1=0 )$ which can have at most 5 real roots or at least 1. (think why ). Geometrically it is the point where the three curves $f(x) , f{-1}(x)$ and x=y do meet together. To check whether f(x) has 5 ,3 or 1 real root we proceed as: $f(x)={ x }^{ 5 }-x+1$
$\frac { d }{ dx } f(x)=5{ x }^{ 4 }-1$ so , $\frac { d }{ dx } f(x)=0$ at $x=\pm \frac { 1 }{ { 5 }^{ 1/4 } }$ but we check for both values of x , f(x)>0. before and after the value range of $x\epsilon (-\frac { 1 }{ { 5 }^{ 1/4 } } ,\frac { 1 }{ { 5 }^{ 1/4 } } )$ it is monotonically increasing or decreasing and prior to the decreasing range there must be a root but for $x\quad >-\frac { 1 }{ { 5 }^{ 1/4 } }$ as we see it is always positive. Therefore it ends to have a singe root and further only a single solution for the equation.

$f(x)=f^{-1}(x)$ can be solved by graphing. Where the graph of $f(x)$ is known and its inverse function will be the reflection of $f$ with $y=x$ . without graphing we can realize that there is only one intersection with $y=x$ and therefore with $f^{-1}$

This problem takes a few steps. First, find f^{-1}(x). To do so, take the equation y = x^5 + 1, switch y and x, then solve for y. Set that equation equal to the original one, and then solve for x. I put it into Wolfram Alpha once it got to that point, in all honesty. It turns out there is just 1 real solution.