if $f(x) = x-\frac{1}{x}.$ Then no. of solution of the equation $f(f(f(x))) = 1$ If $\displaystyle f(x) = x-\frac{1}{x}.$ Then no. of solution of the equation $f(f(f(x))) = 1$
$\underline{\bf{My\;\; Try}}::$ Given $\displaystyle f(x) = x-\frac{1}{x} = \frac{x^2-1}{x}.$ Now Replace $\displaystyle x\rightarrow \frac{1}{x}\;,$ We Get.
$\displaystyle f(f(x)) = x-\frac{1}{x}-\frac{x}{x^2-1} = \frac{x^2-1}{x}-\frac{x}{x^2-1} = \frac{(x^2-1)^2-x^2}{x.(x^2-1)} = \frac{x^4-3x^2+1}{x.(x^2-1)}$
again Replace $\displaystyle x\rightarrow \frac{1}{x}\;,$ We Get
$\displaystyle f(f(f(x))) = \frac{\left(\frac{x^2-1}{x}\right)^4-3\left(\frac{x^2-1}{x}\right)^2+1}{\left(\frac{x^2-1}{x}\right).\left\{\left(\frac{x^2-1}{x}\right)^2-1\right\}}$
Now I did not understand how can I solve This $8^{th}$ Degree equation,
So Help Required,
Thanks
 A: Hint: plot the graph of $f$, and notice that:


*

*it steadily grows from $-\infty$ to $+\infty$ as $x$ grows grom $-\infty$ to $0$

*$x=0$ is a vertical asymptote

*it steadily grows from $-\infty$ to $+\infty$ as $x$ grows grom $0$ to $+\infty$.


Hence $x_1=f(x)$ can be considered as an independent variable on each interval $(-\infty,0)$ and $(0,+\infty)$. It follows that $f \circ f$ has three vertical asymptotes: $x=-1$ and $x=1$ (coming from $f(x)=0$) and $x=0$.
You repeat this argument and discover that $f \circ f\circ f$ has seven vertical asymptotes. Before the smallest asymptote and after the largest one, $f \circ f \circ f$ is steadily increasing from $-\infty$ to $+\infty$. 
Since it is easily shown that $f \circ f \circ f$ is steadily increasing between any two consecutive asymptotes, you conclude that $f(f(f(x)))=1$ has exactly eight distinct solutions.
A: The following method may be unnecessarily time intensive, but will definitely work. Call $f(x) = y$ and $f(y) = f(f(x)) = z$. The equality you need to solve is $f(z) = 1$. This gives a quadratic equation for $z$. Solve it for $z$; this gives you two answers. For these two answers $z_1, z_2$, solve $f(y) = z_1, z_2$, giving you at most 4 answers. Repeat, to solve $f(x) = y_1, ... ,y_4$, giving you at most 8 answers. Eliminate duplicate answers.
A: You want to solve $f(f(f(x)))=1$ for $x$. You can solve it by solving three equations in turn: Set $a:=f(f(x))$, then find $a$ via $f(a)=1$. Then set $b:=f(x)$ and find $b$ via $f(b)=a$. Lastly, find $x$ via $f(x)=b$. Notice also that because $f(-\frac{1}{x_0})=f(x_0)$, for each solution $x_0$, the number $-\frac{1}{x_0}$ is a solution too.
There are $8$ solutions and they all lie in the interval $(-1.9,2.5)$. You can e.g. take Wolfram alpha and plot your ugly rational function and see how often it takes the value $1$. Solving the equation is a one-liner in a program like Mathematica. One solution is $x=-0.40065\dots$, convince yourself by plugging in $-0.4$ and you find $f(f(f(-0.4)))=1.00797\dots$. 
A: Hint: $f^{(-1)}(x)=\frac{x\pm\sqrt{x^2+4}}{2}$, thus $f^{(-1)}(1)=(1\pm\sqrt{5})/2$ and so on.
A: $f(f(x)) =f^{-1}(1)$
$$f(x) = x-\frac{1}{x}$$
$$f^{-1}(x) = \frac{x+\sqrt{x^2+4}}{2}$$
or
$$f^{-1}(x) = \frac{x-\sqrt{x^2+4}}{2}$$
It means we have 2 possible values of $f^{-1}(1)$ 
$f^{-1}(1)=\phi_1=\frac{1+\sqrt{5}}{2}$ or $f^{-1}(1)=\phi_2=\frac{1-\sqrt{5}}{2}$
For $\phi_1=\frac{1+\sqrt{5}}{2}$
$f(f(x)) =\phi_1=\frac{1+\sqrt{5}}{2}$
As you pointed in question we know that $f(f(x)) = x-\frac{1}{x}-\frac{1}{x-\frac{1}{x}} $
$x-\frac{1}{x}=P$  then
$f(f(x)) = P-\cfrac{1}{P}= \phi_1$
You can solve $P$
$P^2-\phi_1P-1=0$
You will find 2 values from here and you will use in it
$x-\frac{1}{x}=P_1$  and $x-\frac{1}{x}=P_2 $ then you will find 4 different x values.
Then you will do the same things for  For $\phi_2=\frac{1-\sqrt{5}}{2}$,
$f(f(x)) = P-\cfrac{1}{P}= \phi_2$ and you will find other 4 different x values
Totally 8 different x , you will find.
