Given as
f(x) = √(x + 1) and g(x) = √(9 - x2)
As we know the square of a real number is never negative.
Therefore, f(x) takes real values only when x + 1 ≥ 0
x ≥ –1, x ∈ [–1, ∞)
The domain of f = [–1, ∞)
Similarly, g(x) takes real values only when 9 – x2 ≥ 0
9 ≥ x2
x2 ≤ 9
x2 – 9 ≤ 0
x2 – 32 ≤ 0
(x + 3)(x – 3) ≤ 0
x ≥ –3 and x ≤ 3
∴ x ∈ [–3, 3]
The domain of g = [–3, 3]
(i) f + g
As we know, (f + g)(x) = f(x) + g(x)
(f + g) (x) = √(x + 1) + √(9 - x2)
The domain of f + g = Domain of f ∩ Domain of g
= [–1, ∞) ∩ [–3, 3]
= [–1, 3]
∴ f + g: [–1, 3] → R is given by (f + g) (x) = f(x) + g(x) = √(x+1) + √(9-x2)
(ii) g – f
As we know, (g – f)(x) = g(x) – f(x)
(g – f) (x) = √(9 - x2) – √(x + 1)
The domain of g – f = Domain of g ∩ Domain of f
= [–3, 3] ∩ [–1, ∞)
= [–1, 3]
∴ g – f: [–1, 3] → R is given by (g – f) (x) = g(x) – f(x) = √(9 - x2) – √(x + 1)
(iii) fg
As we know, (fg) (x) = f(x)g(x)
(fg) (x) = √(x + 1)√(9 - x2)
= √[(x + 1) (9 - x2)]
= √[x(9 - x2) + (9 - x2)]
= √(9x - x3 + 9 - x2)
= √(9 + 9x - x2 - x3)
The domain of fg = Domain of f ∩ Domain of g
= [–1, ∞) ∩ [–3, 3]
= [–1, 3]
∴ fg: [–1, 3] → R is given by (fg) (x) = f(x) g(x) = √(x + 1)√(9 - x2) = √(9 + 9x - x2 - x3)
(iv) f/g
As we know, (f/g) (x) = f(x)/g(x)
(f/g) (x) = √(x + 1)/√(9 - x2)
= √[(x + 1)/(9 - x2)]
The domain of f/g = Domain of f ∩ Domain of g
= [–1, ∞) ∩ [–3, 3]
= [–1, 3]
However, (f/g)(x) is defined for all real values of x ∈ [–1, 3], except for the case when 9 – x2 = 0 or x = ± 3
When x = ±3, (f/g) (x) will be undefined as the division result will be indeterminate.
The domain of f/g = [–1, 3] – {–3, 3}
The domain of f/g = [–1, 3)
∴ f/g: [–1, 3) → R is given by (f/g) (x) = f(x)/g(x) = √(x + 1) / √(9 - x2)
(v) g/f
As we know, (g/f) (x) = g(x)/f(x)
(g/f) (x) = √(9 - x2)/√(x + 1)
= √[(9 - x2)/(x + 1)]
The 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
If 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
As 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)
The 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)
(vii) f2 + 7f
As we know, (f2 + 7f) (x) = f2(x) + (7f)(x)
(f2 + 7f) (x) = f(x) f(x) + 7f(x)
= √(x + 1)√(x + 1) + 7√(x + 1)
= x + 1 + 7√(x + 1)
Domain of f2 + 7f is same as domain of f.
Domain of f2 + 7f = [–1, ∞)
∴ f2 + 7f: [–1, ∞) → R is given by (f2 + 7f) (x) = f(x) f(x) + 7f(x) = x + 1 + 7√(x + 1)
(viii) 5/g
As we know, (5/g) (x) = 5/g(x)
(5/g) (x) = 5/√(9 - x2)
The domain of 5/g = Domain of g = [–3, 3]
However, (5/g) (x) is defined for all real values of x ∈ [–3, 3], except for the case when 9 – x2 = 0 or x = ± 3
If x = ±3, (5/g) (x) will be undefined as the division result will be indeterminate.
The domain of 5/g = [–3, 3] – {–3, 3}
= (–3, 3)
∴ 5/g: (–3, 3) → R is given by (5/g) (x) = 5/g(x) = 5/√(9 - x2)