Solutions:
(v) g/f
We know, (g/f) (x) = g(x)/f(x)
(g/f) (x) = √(9-x2) / √(x+1)
= √[(9-x2) / (x+1)]
Domain of g/f = Domain of f ∩ Domain of g
= [–1, ∞) ∩ [–3, 3]
= [–1, 3]
However, (g/f) (x) is defined for all real values of x ∈ [–1, 3], except for the case when x + 1 = 0 or x = –1
When x = –1, (g/f) (x) will be undefined as the division result will be indeterminate.
Domain of g/f = [–1, 3] – {–1}
Domain of g/f = (–1, 3]
∴ g/f: (–1, 3] → R is given by (g/f) (x) = g(x)/f(x) = √(9-x2) / √(x+1)
(vi) 2f – √5g
We know, (2f – √5g) (x) = 2f(x) – √5g(x)
(2f – √5g) (x) = 2f (x) – √5g (x)
= 2√(x+1) – √5√(9-x2)
= 2√(x+1) – √(45- 5x2)
Domain of 2f – √5g = Domain of f ∩ Domain of g
= [–1, ∞) ∩ [–3, 3]
= [–1, 3]
∴ 2f – √5g: [–1, 3] → R is given by (2f – √5g) (x) = 2f (x) – √5g (x) = 2√(x+1) – √(45- 5x2)