Question

If $f$ is an invertible function such that $f\left(x\right)=x^3+e^{x/2}$ and $g\left(x\right)=f^{-1}\left(x\right)\forall x,$ then the value of $g^{\prime }\left(1\right)$ is

Answer 1

$g\left(x\right)=f^{-1}\left(x\right)$
$\Rightarrow g$ and $f$ are inverse of each other.
$g\left(f\right(x\left)\right)=f^{-1}\left(f\right(x\left)\right)=x$
Differentiating both sides w.r.t. $x,$
$g^{\prime }\left(f\right(x\left)\right)\cdot f^{\prime }\left(x\right)=1$
$\Rightarrow g^{\prime }\left(f\right(x\left)\right)=\frac{1}{f^{\prime }\left(x\right)}=\frac{1}{3x^2+\frac{1}{2}{e}^{x/2}}$
Put $x=0,$ we get
$g^{\prime }\left(1\right)=\frac{1}{f^{\prime }\left(0\right)}=\frac{1}{\frac{1}{2}}=2$