<s>[INST] 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
 [/INST] 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.
</s>