Question -
Answer -
Given:
f(x) = √(x+1) and g(x) = √(9-x2)
We know the square of a real number is never negative.
So, f(x) takes real values only when x + 1 ≥ 0
x ≥ –1, x ∈ [–1, ∞)
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]
Domain of g = [–3, 3]
(i) f+ g
We know, (f + g)(x) = f(x) + g(x)
(f + g) (x) = √(x+1) + √(9-x2)
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
We know, (g – f)(x) = g(x) – f(x)
(g – f) (x) = √(9-x2) –√(x+1)
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
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)
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
We know, (f/g) (x) = f(x)/g(x)
(f/g) (x) = √(x+1) / √(9-x2)
= √[(x+1) / (9-x2)]
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 resultwill be indeterminate.
Domain of f/g = [–1, 3] – {–3, 3}
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
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 resultwill 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)
(vii) f2 + 7f
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 issame 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
We know, (5/g) (x) = 5/g(x)
(5/g) (x) = 5/√(9-x2)
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
When x = ±3, (5/g) (x) will be undefined as the division resultwill be indeterminate.
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)