Question

Let $f : R ^{+} \rightarrow(-7, \infty)$ be a function defined by $f(x)=\frac{ x ^3-7}{ x ^2+1}$ and $g$ be the inverse of $f$ such that $\frac{1}{\alpha} g \left(\frac{1}{\alpha^2+1}\right)=1$ for some $\alpha$. If $k =\frac{ g (\alpha+1)}{ g ( g (\alpha))}$ then find the value of $k$.

Answer 1

$g \left(\frac{1}{\alpha^2+1}\right)=\alpha \Rightarrow f (\alpha)=\frac{1}{\alpha^2+1} \Rightarrow \alpha=2$
$k =\frac{ g (\alpha+1)}{ g ( g (\alpha))}=\frac{ g (3)}{ g ( g (2))}=\frac{ g (3)}{ g (3)}=1$