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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

 

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