finding the inverse of a function

Find the value of parameter $\alpha$ for which the funciion $f(x)=1+\alpha x,\alpha\neq0$ is the inverse of itself. SOLUTION Clearly, $f(x)$ is a bijection from R to itself. Now,  $fof^{-1}(x)=x$ $\Rightarrow f(f^{-1}(x))=x$  $\Rightarrow1+\alpha f^{-1}(x)=x$ $\Rightarrow f^{-1}(x)=\frac{x-1}{\alpha}$ It is given that  $f(x)=f^{-1}(x)$ for all$x\in R$   $\Rightarrow1+{\displaystyle \alpha x=\frac{x-1}{\alpha}}for\: all\: x\in R$   $\Rightarrow{\displaystyle \alpha x+1=(\frac{1}{\alpha})x+\left(\frac{-1}{\alpha}\right)}for\: all\: x\in R$   $\Rightarrow{\displaystyle \alpha=\frac{1}{\alpha}}and1=-{\displaystyle \frac{1}{\alpha}}$   $\Rightarrow\alpha^{2}=1\mathrm{and}\alpha=-1$   $\Rightarrow\alpha=-1.$