If f : R → R be defined as f(x ) = 2x + 3, then f^–1(x) =
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f(x)=y=2x+3
g(y)=x= (y+3)/2
....
g o f(x)= (2x+3-3)/2=x
Thus,
g o f(x)=I x
Similarly,
f o g(y)= I y.
..
Thus,f-¹(x)= x
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