Let A = {1, 2, 3, 4}; B = {3, 5, 7, 9}; C = {7, 23, 47, 79} and

Solution:

$f(x)=2 x+1$

$\Rightarrow f=\{(1,2(1)+1),(2,2(2)+1),(3,2(3)+1),(4,2(4)+1)\}=\{(1,3),(2,5),(3,7),(4,9)\}$

$g(x)=x^{2}-2$

$\Rightarrow g=\left\{\left(3,3^{2}-2\right),\left(5,5^{2}-2\right),\left(7,7^{2}-2\right),\left(9,9^{2}-2\right)\right\}=\{(3,7),(5,23),(7,47),(9,79)\}$

Clearly $f$ and $g$ are bijections and, hence, $f^{-1}: B \rightarrow A$ and $g^{-1}: C \rightarrow B$ exist.

So, $f^{-1}=\{(3,1),(5,2),(7,3),(9,4)\}$

and $g^{-1}=\{(7,3),(23,5),(47,7),(79,9)\}$

Now, $\left(f^{-1} \circ g^{-1}\right): C \rightarrow A$

$f^{-1} o g^{-1}=\{(7,1),(23,2),(47,3),(79,4)\}$$\ldots(1)$

Also, $f: A \rightarrow B$ and $g: B \rightarrow C$,

$\Rightarrow g o f: A \rightarrow C,(g o f)^{-1}: C \rightarrow A$

So, $f^{-1} o g^{-1}$ and $(g o f)^{-1}$ have same domains.

$(g o f)(x)=g(f(x))=g(2 x+1)=(2 x+1)^{2}-2$

$\Rightarrow(g o f)(x)=4 x^{2}+4 x+1-2$

$\Rightarrow(g o f)(x)=4 x^{2}+4 x-1$

Then, $(g o f)(1)=g(f(1))=4+4-1=7$,

$(g o f)(2)=g(f(2))=4+4-1=23$

$(g o f)(3)=g(f(3))=4+4-1=47$ and

$(g o f)(4)=g(f(4))=4+4-1=79$

So, $g o f=\{(1,7),(2,23),(3,47),(4,79)\}$

From $(1)$ and $(2)$, we get:

$(g o f)^{-1}=f^{-1} o g^{-1}$