Given, and h(x) = 2x³ – 3

We know the square of a real number is never negative.

Clearly, f(x) takes real values only when x + 1 ≥ 0

⇒ x ≥ –1

∴ x ∈ [–1, ∞)

Thus, domain of f = [–1, ∞)

g(x) is defined for all real values of x, except for 0.

Thus, domain of g = R – {0}

h(x) is defined for all real values of x.

Thus, domain of h = R

We know (2f + g – h)(x) = (2f)(x) + g(x) – h(x)

⇒ (2f + g – h)(x) = 2f(x) + g(x) – h(x)

Domain of 2f + g – h = Domain of f ∩ Domain of g ∩ Domain of h

⇒ Domain of 2f + g – h = [–1, ∞) ∩ R – {0} ∩ R

∴ Domain of 2f + g – h = [–1, ∞) – {0}

i. (2f + g – h)(1)

We have

ii. (2f + g – h)(0)

0 is not in the domain of (2f + g – h)(x).

Hence, (2f + g – h)(0) does not exist.

Thus, and (2f + g – h)(0) does not exist as 0 is not in the domain of (2f + g – h)(x).