Question

Let $f\left(x\right)=2x^3-x$ and $g$ is the inverse of function $f.$ Then the value of $4-125g^{\prime \prime }\left(1\right)$ is

Answer 1

Given : $f\left(x\right)=2x^3-x\Rightarrow f\left(1\right)=1$
Differentiating both sides w.r.t. $x$
$f^{\prime }\left(x\right)=6x^2-1\Rightarrow f^{\prime }\left(1\right)=5$
Differentiating both sides w.r.t. $x$
$f^{\prime \prime }\left(x\right)=12x\Rightarrow f^{\prime \prime }\left(1\right)=12$
and $f^{-1}\left(x\right)=g\left(x\right)\Rightarrow g^{-1}\left(x\right)=f\left(x\right)$
$\therefore g\left(f\right(x\left)\right)=x$
Differentiating both sides w.r.t. $x$
$g^{\prime }\left(f\right(x\left)\right)\cdot f^{\prime }\left(x\right)=1$
$\Rightarrow g^{\prime }\left(f\right(x\left)\right)=\frac{1}{f^{\prime }\left(x\right)}$
Differentiating both sides w.r.t. $x$
$\Rightarrow g^{\prime \prime }\left(f\right(x\left)\right)\cdot f^{\prime }\left(x\right)=-\frac{f^{\prime \prime }\left(x\right)}{\left[f^{\prime }\right(x){]}^2}$
$\Rightarrow g^{\prime \prime }\left(f\right(x\left)\right)=-\frac{f^{\prime \prime }\left(x\right)}{\left[f^{\prime }\right(x){]}^3}$
$\Rightarrow g^{\prime \prime }\left(f\right(1\left)\right)=-\frac{f^{\prime \prime }\left(1\right)}{\left[f^{\prime }\right(1){]}^3}$
$\Rightarrow g^{\prime \prime }\left(1\right)=-\frac{12}{5^3}=\frac{-12}{125}$
$\therefore 4-125g^{\prime \prime }\left(1\right)=16$