Q. 1 B5.0( 3 Votes )

# Find f + g, f – g

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)

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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