Verified answer

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f(x) = x - 2

g(x) = √ ̅x̅  + 1

So,

f(g(x))  =   (√ ̅x̅  + 1) - 2   =   √ ̅x̅  - 1        ← this is real only if the

                                                             square root is real so

                                                             that x must be non negative.

           ANSWER

Domain of f(g(x)) = {x│ x ≥ 0 }

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Dom f = all reals; Dom g = {x | x ≥ 0}

So Dom f◦g = Dom f ∩ Dom g = {x | x ≥ 0}

To check that this is true, f(g(x) = g(x) – 2 = √x + 1 – 2 = √x – 1

Since negatative are not allowed under square roots (assuming real functions, not complex), it should now be clear.

f(x) = x - 2

f(g(x)) = √x + 1 -2

=> f(g(x)) = √x -1

The domain of f(g(x)) is the set of all non negative real numbers.

√x -1.

All real numbers.

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