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)