Given,

f(x) = x + 1 and 

g(x) = 2x – 3 

Clearly,

Both f(x) and g(x) exist for all real values of x. 

Hence,

Domain of f = Domain of g = R 

Range of f = Range of g = R 

i. f + g 

We know,

(f + g)(x) = f(x) + g(x) 

⇒ (f + g)(x) = x + 1 + 2x – 3 

∴ (f + g)(x) = 3x – 2 

Domain of f + g = Domain of f ∩ Domain of g 

⇒ Domain of f + g = R ∩ R 

∴ Domain of f + g = R 

Thus,

f + g : R → R is given by 

(f + g)(x) = 3x – 2 

ii. f – g 

We know 

(f – g)(x) = f(x) – g(x) 

⇒ (f – g)(x) = x + 1 – (2x – 3) 

⇒ (f – g)(x) = x + 1 – 2x + 3 

∴ (f – g)(x) = –x + 4 

Domain of f – g = Domain of f ∩ Domain of g 

⇒ Domain of f – g = R ∩ R 

∴ Domain of f – g = R 

Thus, 

f – g : R → R is given by 

(f – g)(x) = –x + 4

iii. \(\frac{f}{g}\) 

We know,

(\(\frac{f}{g}\))(x) = \(\frac{f(x)}{g(x)}\)

∴ (\(\frac{f}{g}\))(x) = \(\frac{x+1}{2x-3}\) 

Clearly,

(\(\frac{f}{g}\))(x) is defined for all real values of x,

Except for the case when 2x – 3 = 0 or x = \(\frac{3}{2}\).

When  x = \(\frac{3}{2}\),(\(\frac{f}{g}\))(x) will be undefined as the division result will be indeterminate. 

Thus,

Domain of \(\frac{f}{g}\) = R – \(\{\frac{3}{2}\}\)

Thus, 

\(\frac{f}{g}\) : R – \(\{\frac{3}{2}\}\)  → R is given by

(\(\frac{f}{g}\))(x) = \(\frac{x+1}{2x-3}\)