Question

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

Answer 1

Answer: 1

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