Given f(x) = 2x + 5 and g(x) = x² + x

Clearly, both f(x) and g(x) are defined for all x ∈ R.

Hence, domain of f = domain of g = R

i. f + g

We know (f + g)(x) = f(x) + g(x)

⇒ (f + g)(x) = 2x + 5 + x² + x

∴ (f + g)(x) = x² + 3x + 5

Clearly, (f + g)(x) is defined for all real numbers x.

∴ The domain of (f + g) is R

ii. f – g

We know (f – g)(x) = f(x) – g(x)

⇒ (f – g)(x) = 2x + 5 – (x² + x)

⇒ (f – g)(x) = 2x + 5 – x² – x

∴ (f – g)(x) = 5 + x – x²

Clearly, (f – g)(x) is defined for all real numbers x.

∴ The domain of (f – g) is R

iii. fg

We know (fg)(x) = f(x)g(x)

⇒ (fg)(x) = (2x + 5)(x² + x)

⇒ (fg)(x) = 2x(x² + x) + 5(x² + x)

⇒ (fg)(x) = 2x³ + 2x² + 5x² + 5x

∴ (fg)(x) = 2x³ + 7x² + 5x

Clearly, (fg)(x) is defined for all real numbers x.

∴ The domain of fg is R

iv.

We know

Clearly, is defined for all real values of x, except for the case when x² + x = 0.

x² + x = 0

⇒ x(x + 1) = 0

⇒ x = 0 or x + 1 = 0

⇒ x = 0 or –1

When x = 0 or –1, will be undefined as the division result will be indeterminate.

Thus, domain of = R – {–1, 0}