Question

Let $f\left(x\right)=e^{x^3+x}$ and $g\left(x\right)=f^{-1}\left(x\right)$ then |$g^{\prime \prime }\left(1\right)$| is

Answer 1

$f\left(x\right)=e^{x^3+x}$
$f^{\prime }\left(x\right)=e^{x^3+x}\times (3x^2+1)$
$\because g\left(x\right)=f^{-1}\left(x\right)\Rightarrow f\left(g\right(x\left)\right)=x$
$\Rightarrow f^{\prime }\left(g\right(x\left)\right)\times g^{\prime }\left(x\right)=1$
$g^{\prime }\left(x\right)=\frac{1}{f^{\prime }\left(g\right(x\left)\right)}$
$\because f\left(g\right(x\left)\right)=x\Rightarrow f\left(g\right(1\left)\right)=1\Rightarrow g\left(1\right)=0$
$Now,g^{\prime }\left(1\right)=\frac{1}{f^{\prime }\left(g\right(1\left)\right)}=\frac{1}{f^{\prime }\left(0\right)}=1$
$g^{\prime \prime }\left(x\right)=-\frac{1}{\left(f^{\prime }\right(g\left(x\right)){)}^2}\times f^{\prime \prime }\left(g\right(x\left)\right)\times g^{\prime }\left(x\right)$
$g^{\prime \prime }\left(1\right)=-\frac{1}{(1{)}^2}\times f^{\prime \prime }\left(0\right)\times 1$
$g^{\prime \prime }\left(1\right)=-1$