Given as
f(x) = 2x + 5 and g(x) = x2 + x
Here, both f(x) and g(x) are defined for all x ∈ R.
Therefore, domain of f = domain of g = R
(i) f + g
As we know, (f + g)(x) = f(x) + g(x)
(f + g)(x) = 2x + 5 + x2 + x
= x2 + 3x + 5
Now, (f + g)(x) is defined for all real numbers x.
∴ The domain of (f + g) is R
(ii) f – g
As we know, (f – g)(x) = f(x) – g(x)
(f – g)(x) = 2x + 5 – (x2 + x)
= 2x + 5 – x2 – x
= 5 + x – x2
(f – g)(x) is defined for all real numbers x.
∴ The domain of (f – g) is R
(iii) fg
As we know, (fg)(x) = f(x)g(x)
(fg)(x) = (2x + 5)(x2 + x)
= 2x(x2 + x) + 5(x2 + x)
= 2x3 + 2x2 + 5x2 + 5x
= 2x3 + 7x2 + 5x
(fg)(x) is defined for all real numbers x.
∴ The domain of fg is R
(iv) f/g
As we know, (f/g) (x) = f(x)/g(x)
(f/g) (x) = (2x+5)/(x2+x)
Here, (f/g) (x) 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, (f/g) (x) will be undefined as the division result will be indeterminate.
Thus, the domain of f/g = R – {–1, 0}