Question

Let $y=f\left(x\right)$ is an invertible function satisfying $f\left(1\right)=5,f^{'}\left(1\right)=2,f^{"}\left(1\right)=4,$ then the absolute value of $2.\left(f^{- 1}\right)^{"}\left(5\right)$ is equal to

Answer 1

$y=f\left(x\right)\Rightarrow f\left(1\right)=5\Rightarrow y=5$ when $x=1$
$x=g\left(y\right)g\left(y\right) \rightarrow $ inverse of $f\left(x\right)$
$1=g^{'}\left(y\right)\cdot f^{'}\left(x\right)\Rightarrow g^{"}\left(y\right)=-\frac{f^{"} \left(x\right)}{\left(f^{'} \left(x\right)\right)^{3}}$
$\Rightarrow g^{"}\left(5\right)=\frac{- 4}{8}=$ (Put $x=1,y=5$ )