How come the function and the inverse of the function are the same? What is the inverse of the function:
$$f(x)=\frac{x+2}{5x-1}$$
?
Answer:
$$f^{-1}(x)=\frac{x+2}{5x-1}$$
Can one of you explain how the inverse is the same exact thing as the original equation?
 A: Let $f(x)=y$. So we have $y=\frac{x+2}{5x-1}$. Now "swap" the variables so we have 
$$x = \frac{y'+2}{5y'-1}$$
where $y' = f^{-1}(x)$ (to distinguish $y'$ from $y$). Now let's solve for $y'$.
\begin{align}
x &= \frac{y'+2}{5y'-1} \\
x(5y'-1) &= y'+2 & \text{multiply both sides by } 5y'-1 \\
5xy' - x&=y'+2 & \text{distribute $x$ on the left side} \\ 
5xy'-y'-x&=2 & \text{move all $y'$ to left side} \\
5xy'-y'&=x+2 & \text{and add $x$ to right side} \\
(5x-1)y'&=x+2 & \text{factor out $y'$ on left side} \\
y'&=\frac{x+2}{5x-1} & \text{divide both sides by $5x-1$}
\end{align}
Thus, $f^{-1}(x)=y'=\frac{x+2}{5x-1}=y=f(x)$.
A: The inverse is not in general "the same exact thing as the original equation".
Generally, $f(x)\ne f^{-1}(x)$, but this is not always true. For example, consider the function $f(x) = -x$. This function is just the function that negates its input. Of course, if your negate your input twice, you get the original input. Put another way, to reverse the operation of negating your input, you simply negate it again.
It just happens to be the case that your function satisfies the same property. Namely,
$$f(f(x)) = \frac{f(x)+2}{5f(x)-1} = \frac{\frac{x+2}{5x-1}+2}{5\frac{x+2}{5x-1} - 1} = x$$
and hence $f(x)$ is its own inverse. (ie, to reverse $f(x)$, simply apply it again!)
A: Let $y=\frac{x+2}{5x-1}$. Then $5xy-y=x+2$, so $x(5y-1)=5xy-x=2+y$, so $x=\frac{2+y}{5y-1}$.
A: First, note that this is not always the case.
As a geometric intuition, more than anything, consider the following:
the inverse of a function, $f^{-1}(x)$, graphically, is a reflection of $y=f(x)$in the line $y=x$ (note that a function has an inverse iff it is injective).
So, with regards to this statement, we should ask ourselves:"Under what circumstances is a function self-inverse (i.e. its own inverse)"?
The answer is: a function $f(x)=f^{-1}(x) \iff$ the graph $y=f(x)$ is symmetric about the line $y=x$, which is the case in this question, as we can see from this graph:

A: First we have
$$f'(x)=-\frac{11}{(5x-1)^2}$$
hence $f$ is strictly decreasing on every interval $(-\infty,\frac15)$ and $(\frac15,+\infty)$ so its restriction has an inverse function on these intervals.
Now it suffices to verify that $f(f(x))=x$:
$$\frac{f(x)+2}{5f(x)-1}=\frac{\frac{x+2}{5x-1}+2}{5\frac{x+2}{5x-1}-1}=\frac{x+2+10x-2}{5x+10-5x+1}=x$$
and geometrically this means that the curve of $f$ is symmetric relative to the line with equation $y=x$.
A: $$f(x) = \frac{x+2}{5x-1}$$
$$(5x-1)f(x)=x+2$$
$$5xf(x)-f(x)=x+2$$
$$5xf(x)-x=f(x)+2$$
$$x(5f(x)-1)=f(x)+2$$
$$x=\frac{f(x)+2}{5f(x)-1}$$
Functions which are there own inverses are called involutions.
A: Write it as: $$(5x-1)y=x+2.\tag{$5x-1\neq0$}$$
Now replace all $x$s with $y\,$s: $$(5y-1)x=y+2.$$
Solving for $y$ gives rise to: $$y=\dfrac{x+2}{5x-1}.\tag{$5x-1\neq0$}$$
Which happens to be our inverse function.
