(i) f(x) = (2x + 1)/(x - 9)
If the denominator vanishes when x = 9
So f(x) is not defined at x = 9
∴ Domain is x ∈ [R – {9}]
(ii) p(x) = -5/(4x2 + 1)
p(x) is defined for all values of x. So domain is x ∈ R.
(iii) g(x) = √(x - 2)
When x < 2 g(x) becomes complex. But given “g” is real valued function.
So x > 2
Domain x ∈ (2, α)
(iv) h (x) = x + 6
For all values of x, h(x) is defined. Hence domain is x ∈ R.