If \(f(x)=\frac{4 x+7}{7 x-4}\), then the value of \(f\{f[f(2)]\}=\)

\(f(x)=\frac{4 x+7}{7 x-4}\)
\(f(2)=\frac{8+7}{14-4}=\frac{15}{10}=\frac{3}{2}\)
\(f[f(2)]=f\left(\frac{3}{2}\right)=\frac{\left(4 \times \frac{3}{2}\right)+7}{\left(7 \times \frac{3}{2}\right)-4}=\frac{6+7}{\left(\frac{21-8}{2}\right)}=\frac{13 \times 2}{13}=2\)
\(f\{f[f(2)]\}=f(2)=\frac{3}{2}\)
This problem can also be solved as follows :
\(\begin{aligned}
\mathrm{f}(\mathrm{x}) \quad=\frac{4 \mathrm{x}+7}{7 \mathrm{x}-4} & \Rightarrow \mathrm{f}[\mathrm{f}(\mathrm{x})]=\mathrm{f}\left[\frac{4 \mathrm{x}+7}{7 \mathrm{x}-4}\right] \\
\therefore \mathrm{f}\left[\frac{4 \mathrm{x}+7}{7 \mathrm{x}-4}\right] \quad &=\frac{4\left(\frac{4 \mathrm{x}+7}{7 \mathrm{x}-4}\right)+7}{7\left(\frac{4 \mathrm{x}+7}{7 \mathrm{x}-4}\right)-4} \\
&=\frac{16 \mathrm{x}+28+49 \mathrm{x}-28}{28 \mathrm{x}+49-28 \mathrm{x}+16}=\frac{65 \mathrm{x}}{65}=\mathrm{x} \\
\therefore \quad \mathrm{f}\{\mathrm{f}[\mathrm{f}(\mathrm{x})]\} &=\mathrm{f}\{\mathrm{x}\}=\frac{4 \mathrm{x}+7}{7 \mathrm{x}-4} \\
\therefore \quad \mathrm{f}\{\mathrm{f}[\mathrm{f}(2)]\} &=\frac{4(2)+7}{7(2)-4}=\frac{15}{10}=\frac{3}{2}
\end{aligned}\)