Let f(x) =
Question:

Let $f(x)=\frac{\alpha x}{x+1}, x \neq-1$. Then write the value of a satisfying $f(f(x))=x$ for all $x \neq-1$.

Solution:

Given:

$f(x)=\frac{\alpha x}{x+1}, x \neq-1$

Since $f(f(x))=x$

$\frac{\alpha\left(\frac{\alpha x}{x+1}\right)}{\frac{\alpha x}{x+1}+1}=x$

$\Rightarrow \frac{\alpha^{2} x}{\alpha x+x+1}=x$

$\Rightarrow \alpha^{2} x-\alpha x^{2}-\left(x^{2}+x\right)=0$

Solving | the quadratic lequation in $\alpha:$

$\alpha=\frac{x^{2} \pm \sqrt{x^{4}+4 x\left(x^{2}+x\right)}}{2 x}$

$\Rightarrow \alpha=x+1$ or $-1$

Since, $\alpha \neq x+1$

$\alpha=-1$