Question:
Let $f: R \rightarrow R$ be defined as $f(x)=\frac{2 x-3}{4} .$ Write $f o f^{-1}$ (1).
Solution:
Let $f^{-1}(x)=y$ ...(1)
$\Rightarrow f(y)=x$
$\Rightarrow \frac{2 y-3}{4}=x$
$\Rightarrow 2 y-3=4 x$
$\Rightarrow 2 y=4 x+3$
$\Rightarrow y=\frac{4 x+3}{2}$
$\Rightarrow f^{-1}(x)=\frac{4 x+3}{2} \quad[$ from $(1)]$
$\Rightarrow f^{-1}(x)=\frac{4 x+3}{2}$
$\therefore\left(f o f^{-1}\right)(1)=f\left(\frac{4(1)+3}{2}\right)=f\left(\frac{7}{2}\right)=\frac{2\left(\frac{7}{2}\right)-3}{4}=\frac{7-3}{4}=\frac{4}{4}=1$