Find f + g, f – g, cf (c ∈ R, c ≠ 0), fg, 1/f and f/g in each of the following:
and ![]()
and ![]()
We have f(x) : [1, ∞) → R+ and g(x) : [–1, ∞) → R+ as real square root is defined only for non-negative numbers.
(a) f + g
We know (f + g)(x) = f(x) + g(x)
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Domain of f + g = Domain of f ∩ Domain of g
⇒ Domain of f + g = [1, ∞) ∩ [–1, ∞)
∴ Domain of f + g = [1, ∞)
Thus, f + g : [1, ∞) → R is given by ![]()
(b) f – g
We know (f – g)(x) = f(x) – g(x)
![]()
Domain of f – g = Domain of f ∩ Domain of g
⇒ Domain of f – g = [1, ∞) ∩ [–1, ∞)
∴ Domain of f – g = [1, ∞)
Thus, f – g : [1, ∞) → R is given by ![]()
(c) cf (c ∈ R, c ≠ 0)
We know (cf)(x) = c × f(x)
![]()
Domain of cf = Domain of f
∴ Domain of cf = [1, ∞)
Thus, cf : [1, ∞) → R is given by ![]()
(d) fg
We know (fg)(x) = f(x)g(x)
![]()
![]()
Domain of fg = Domain of f ∩ Domain of g
⇒ Domain of fg = [1, ∞) ∩ [–1, ∞)
∴ Domain of fg = [1, ∞)
Thus, fg : [1, ∞) → R is given by ![]()
(e) ![]()
We know ![]()
![]()
Domain of
= Domain of f
∴ Domain of
= [1, ∞)
Observe that
is also undefined when x – 1 = 0 or x = 1.
Thus,
: (1, ∞) → R is given by ![]()
(f) ![]()
We know ![]()


Domain of
= Domain of f ∩ Domain of g
⇒ Domain of
= [1, ∞) ∩ [–1, ∞)
∴ Domain of
= [1, ∞)
Thus,
: [1, ∞) → R is given by ![]()
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