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Question
` if f : (-π/2 , π/2)` → R and g : [−1, 1]→ R be defined as f(x) = tan x and g(x) = `sqrt(1 - x^2)` respectively, describe fog and gof.
Solution
`g (x) = sqrt (1- x^2)`
⇒ x2 ≥ 0, ∀x ∈[−1, 1]
⇒ −x2 ≤ 0, ∀x ∈ [−1, 1]
⇒ 1−x2 ≤ 1, ∀x ∈ [−1, 1]
We know that 1 - x2 ≥0
⇒ 0≤1 -x2≤1
⇒ Range of g(x) = [0, 1]
So, f : ` ( π/2 , π/2)` → R and g : [−1, 1]→ [0, 1]
Computation of fog :
Clearly, the range of g is a subset of the domain of f.
So, fog : [−1, 1] → R
(fog) (x) = f (g (x))
= f `( sqrt (1 - x^2))`
= tan `sqrt (1 - x^2)`
Computation of gof:
Clearly, the range of f is not a subset of the domain of g.
⇒ Domain (gof) = { x ∈ domain of f and f (x)∈domain of g}
⇒ Domain (gof) =`{ x in ((-π)/2 , π/2)` and tan x ∈ [−1,1] }`
⇒ Domain (gof) = `{x in((-π)/2 , π/2) and x in (-π)/4 , π/4 )} `
⇒ Domain (gof) = `{x in ((-x)/4 , π/4) ,}`
Now, gof : `((-x)/4 , π/4)` → R
So, (gof) (x) = g (f (x))
= g (tan x)
= `sqrt(1- tan^2 x)`
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