Let f:(2, ∞) → R: f(x) = \(\sqrt{x-2}\), and g: (2, ∞) → R: g

1.	Let f:(2, ∞) → R: f(x) = \(\sqrt{x-2}\), and g: (2, ∞) → R: g(x) = \(\sqrt{x+2}\) Find: (i) (f + g) (x) (ii) (f - g) (x) (iii) (fg) (x)
Answer» Given f(x) = \(\sqrt{x-2}\): x > and g(x) = \(\sqrt{x+2}\): x > 2 (i) To find: (f + g) (x) Domain(f) = (2, ∞) Range(f) = (0, ∞) Domain(g) = (2, ∞) Range(g) = (2, ∞) (f + g) (x) = f(x) + g(x) = \(\sqrt{x-2}\) + \(\sqrt{x+2}\) Therefore, (f + g) (x) = \(\sqrt{x-2}\) + \(\sqrt{x+2}\) (ii) To find:(f - g)(x) Range(g) ⊆ Domain(f) Therefore, (f - g)(x) exists. (f - g)(x) = f(x) – g(x) = \(\sqrt{x-2}\) + \(\sqrt{x+2}\) Therefore, (f - g) (x) = \(\sqrt{x-2}\) - \(\sqrt{x+2}\) (iii) To find:(fg)(x) (fg)(x) = f(x).g(x) = \(\sqrt{(x-2)}\) .\(\sqrt{(x+2)}\) = \(\sqrt{(x-2)(x+2)}\) = \(\sqrt{(x^2+2^2)}\) ( ∵ a² - b² = (a - b)(a + b)) = \(\sqrt{x^2-4}\) Therefore, (fg)(x) = \(\sqrt{x^2-4}\)

Let f:(2, ∞) → R: f(x) = \(\sqrt{x-2}\), and g: (2, ∞) → R: g(x) = \(\sqrt{x+2}\) Find: (i) (f + g) (x) (ii) (f - g) (x) (iii) (fg) (x)

Answer»

Given

f(x) = \(\sqrt{x-2}\): x > and g(x) = \(\sqrt{x+2}\): x > 2

(i) To find: (f + g) (x)

Domain(f) = (2, ∞)

Range(f) = (0, ∞)

Domain(g) = (2, ∞)

Range(g) = (2, ∞)

(f + g) (x) = f(x) + g(x)

= \(\sqrt{x-2}\) + \(\sqrt{x+2}\)

Therefore,

(f + g) (x) = \(\sqrt{x-2}\) + \(\sqrt{x+2}\)

(ii) To find:(f - g)(x)

Range(g) ⊆ Domain(f)

Therefore,

(f - g)(x) exists.

(f - g)(x) = f(x) – g(x)

= \(\sqrt{x-2}\) + \(\sqrt{x+2}\)

Therefore,

(f - g) (x) = \(\sqrt{x-2}\) - \(\sqrt{x+2}\)

(iii) To find:(fg)(x)

(fg)(x) = f(x).g(x)

= \(\sqrt{(x-2)}\) .\(\sqrt{(x+2)}\)

= \(\sqrt{(x-2)(x+2)}\)

= \(\sqrt{(x^2+2^2)}\) ( ∵ a² - b² = (a - b)(a + b))

= \(\sqrt{x^2-4}\)

Therefore,

(fg)(x) = \(\sqrt{x^2-4}\)