Solution:
We need to Show: fof 
It is known that fof 
![\begin{array}{l} =\mathrm{f}\left(\frac{4 x+3}{6 x-4}\right) \\ =\frac{4 \frac{4 x+3}{6 x-4}+3}{6 \frac{1 x+3}{6 x-4}-4} \quad\left[\because \mathrm{f}(\mathrm{x})=\frac{4 x+3}{6 x-4}\right] \\ =\frac{\frac{4(4 x+3)+3(6 x-4)}{6 x-4}}{\frac{6(4 x+3)-4(6 x-4)}{6 x-4}} \\ =\frac{4(4 x+3)+3(6 x-4)}{6(4 x+3)-4(6 x-4)} \\ =\frac{16 x+12+18 x-13)}{24 x+18-24 x-16} \\ =\frac{34 x}{34} \\ =\mathrm{x} \end{array}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9670b48fda1aee3a23161e4123b82fa2_l3.png "Rendered by QuickLaTeX.com")
As a result, proved that 
It is known that 
![\begin{array}{l} \text { fof }-1(x)=f o f(x) \quad[\because \operatorname{fof}(x)=x)] \\ f(f-1(x))=f(f(x)) \\ f_{1}(x)=\frac{4 x+3}{6 x-4} \end{array}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cf9b6724b769e8937727cffaf9cd26f6_l3.png "Rendered by QuickLaTeX.com")
If 
Solution:
We need to Show: fof
It is known that fof
As a result, proved that
It is known that
