i. f(x) = x³ + 1 and g(x) = x + 1

We have f(x) : R → R and g(x) : R → R

(a) f + g

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

⇒ (f + g)(x) = x³ + 1 + x + 1

∴ (f + g)(x) = x³ + x + 2

Clearly, (f + g)(x) : R → R

Thus, f + g : R → R is given by (f + g)(x) = x³ + x + 2

(b) f – g

We know (f – g)(x) = f(x) – g(x)

⇒ (f – g)(x) = x³ + 1 – (x + 1)

⇒ (f – g)(x) = x³ + 1 – x – 1

∴ (f – g)(x) = x³ – x

Clearly, (f – g)(x) : R → R

Thus, f – g : R → R is given by (f – g)(x) = x³ – x

(c) cf (c ∈ R, c ≠ 0)

We know (cf)(x) = c × f(x)

⇒ (cf)(x) = c(x³ + 1)

∴ (cf)(x) = cx³ + c

Clearly, (cf)(x) : R → R

Thus, cf : R → R is given by (cf)(x) = cx³ + c

(d) fg

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

⇒ (fg)(x) = (x³ + 1)(x + 1)

⇒ (fg)(x) = (x + 1)(x² – x + 1)(x + 1)

∴ (fg)(x) = (x + 1)²(x² – x + 1)

Clearly, (fg)(x) : R → R

Thus, fg : R → R is given by (fg)(x) = (x + 1)²(x² – x + 1)

(e)

We know

Observe that is undefined when f(x) = 0 or when x = – 1.

Thus, : R – {–1} → R is given by

(f)

We know

Observe that is undefined when g(x) = 0 or when x = –1.

Using x³ + 1 = (x + 1)(x² – x + 1), we have

Thus, : R – {–1} → R is given by