Question:
Let $f: R \rightarrow R: f(x)(2 x-3)$ and $g: R \rightarrow R: g(x)=\frac{1}{2}(x+3)$
Show that $(f \circ g)=l_{R}=(g \circ f)$
Solution:
To prove: (f o g) = IR = (g o f).
Formula used: (i) f o g = f(g(x))
(ii) g o f = g(f(x))
Given: (i) $f: R \rightarrow R: f(x)=(2 x-3)$
(ii) $g: R \rightarrow R: g(x)=\frac{1}{2}(x+3)$
Solution: We have,
f o g = f(g(x))
$=f\left(\frac{1}{2}(x+3)\right)$
$=\left[2\left(\frac{1}{2}(x+3)\right)-3\right]$
$=x+3-3$
$=x$
$=I_{R}$
g o f = g(f(x))
$=g(2 x-3)$
$=\frac{1}{2}(2 x-3+3)$
$=\frac{1}{2}(2 x)$
$=x$
$=I_{R}$
Clearly we can see that $(f \circ g)=I_{R}=(g \circ f)=x$
Hence Proved