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Question
Find fog and gof if : f(x) = `x^2` + 2 , g (x) = 1 − `1/ (1-x)`.
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Solution
f (x) = x2+ 2
f : R → [ 2, ∞ )
g (x) = 1 `- 1/(1-x)`
For domain of g : 1− x ≠ 0
⇒ x ≠ 1
⇒ Domain of g = R−{1}
g (x )= `1 - 1/(1-x) = (1-x-1)/(1-x) = (-x)/(1-x)`
For range of g :
`y = (- x)/ (1-x)`
⇒ y − xy = − x
⇒ y = xy − x
⇒ y = x (y−1)
⇒ `x = y/(y-1)`
Range of g =R−{1}
So, g : R−{1}→R−{1}
Computing fog :
Clearly, the range of g is a subset of the domain of f.
⇒ fog : R − {1}→ R
(fog) (x) = f (g (x))
`= f ((-x)/ (x-1) )`
`= ((-x)/ (x-1))^2 + 2`
`=(x^2 + 2x^2 +2-4x)/(1-x)^2`
`= (3x^2-4x +2 )/ (1-x)^2`
Computing gof :
Clearly, the range of f is a subset of the domain of g.
⇒ gof : R→R
(gof) (x) = g (f (x))
= g ( x2 + 2 )
`= 1- 1/(1-(x^2 + 2))`
`= - 1/(1-(x^2 + 2))`
`= (x^2 + 2)/(x^2 + 1)`
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