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Question
if `f (x) = sqrt(1-x)` and g(x) = `log_e` x are two real functions, then describe functions fog and gof.
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Solution
`f (x) = sqrt(1-x)`
For domain, 1-x≥0
⇒ x≤1
⇒ domain of f = (−∞, 1]
⇒ f : (−∞, 1] → (0,∞)
g(x) = loge x
Clearly, g : (0, ∞) → R
Computation of fog :
Clearly, the range of g is not a subset of the domain of f.
So,we need to compute the domain of fog.
⇒ Domain (fog) = {x : x ∈ Domain (g) and g(x) ∈ Domain of f}
⇒ Domain (fog) = {x: x ∈ (0, ∞) and loge x ∈ (−∞, 1]}
⇒ Domain (fog) = { x: x ∈ (0, ∞) and x ∈ (0, e] }
⇒ Domain (fog)= {x : x ∈ (0, e]}
⇒ Domain (fog )= (0, e]
⇒ fog : (0, e) → R
So, (fog) (x) = f (g (x))
= f (loge x)
= `sqrt( 1-log_e x)`
Computation of gof:Clearly, the range of f is a subset of the domain of g.
⇒ gof : (−∞,1] → R
⇒ (gof) (x) = g (f (x))
= `g (sqrt(1-x))`
= `log_e sqrt (1 - x)`
= `log_e (1 - x)^(1/2)`
= `1/2 log_e (1-x)`
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