Let f : R → R be defined as f x = x 3 + x - 5 . If g x is a function such that f g x = x , ∀ x…

Here, $g\left(f\left(x\right)\right)=x$ as $f\left(x\right)$ and $g\left(x\right)$ are inverse of each other.

Now, $g^{\text{'}}\left(f\left(x\right)\right)f^{\text{'}}\left(x\right)=1$

$\Rightarrow g^{\text{'}}\left(f\left(x\right)\right)=\frac{1}{f^{\text{'}}\left(x\right)} \dots \left(i\right)$

Now $f\left(x\right)=63 \Rightarrow x^3+x-5=63$

$\Rightarrow x^3+x-68=0$

So $x=4$ satisfies the above equation

$g^{\text{'}}\left(63\right)=\frac{1}{f^{\text{'}}\left(4\right)}$  from $\left(i\right)$

$=\frac{1}{3{\left(4\right)}^2+1}=\frac{1}{49}$