Let $f: R \rightarrow R$ and $g: R \rightarrow R$ be differentiable functions such that $f(x)=x^3+3 x+2, g(f(x))=x$ for all $x \in R$. Then, $g^{\prime}(2)=$

$\frac{1}{3}$

We have,

∴  $g(f(x))=x$ for all $x \in R$

$\Rightarrow gof(x)=x$ for all $x \in R$

$\Rightarrow g=f^{-1}$ and $g'(f(x)) f'(x)=1$ for all $x \in R$          .......(i)

Now,  $f(x)=2 \Rightarrow x^3+3 x+2=2 \Rightarrow x\left(x^2+3\right)=0 \Rightarrow x=0$

∴  $f(0)=2$

Putting x = 0 in (i). we obtain

$g'(f(0)) f'(0)=1$

$\Rightarrow g'(2) \times f'(0)=1$

$\Rightarrow g'(2)=\frac{1}{f'(0)}=\frac{1}{3}$                $\left[\begin{array}{l} ∵ f(x)=x^3+3 x+2 \\ \Rightarrow f'(x)=3 x^2+3 \Rightarrow f'(0)=3\end{array}\right]$