Solve this

Solution:

Given:

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

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

Domain $(f)=(2, \infty)$

Range $(f)=(0, \infty)$

Domain $(g)=(2, \infty)$

Range $(\mathrm{g})=(2, \infty)$

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

$=\sqrt{x-2}+\sqrt{x+2}$

Therefore,

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

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

Range $(\mathrm{g}) \subseteq$ 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)

$(f g)(x)=f(x) \cdot g(x)$

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

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

$=\sqrt{x^{2}-2^{2}}\left(\because a^{2}-b^{2}=(a-b)(a+b)\right)$

$=\sqrt{x^{2}-4}$

Therefore,

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