Let f and g be two real functions defined by f(x)

Solution:

Given:

$f(x)=\sqrt{x+1}$ and $g(x)=\sqrt{9-x^{2}}$

Clearly, $f(x)=\sqrt{x+1}$ is defined for all $x \geq-1$.

Thus, domain $(f)=[-1, \infty]$

Again,

$g(x)=\sqrt{9-x^{2}}$ is defined for

$9-x^{2} \geq 0 \Rightarrow x^{2}-9 \leq 0$

$\Rightarrow x^{2}-3^{2} \leq 0$

$\Rightarrow(x+3)(x-3) \leq 0$

$\Rightarrow x \in[-3,3]$

Thus, domain $(g)=[-3,3]$

Now,

domain $(f) \cap$ domain $(g)=[-1, \infty] \cap[-3,3]$

$=[-1,3]$

(i) $(f+g):[-1,3] \rightarrow R$ is given by $(f+g)(x)=f(x)+g(x)=\sqrt{x+1}+\sqrt{9-x^{2}}$.

(ii) $(g-f):[-1,3] \rightarrow R$ is given by $(g-f)(x)=g(x)-f(x)=\sqrt{9-x^{2}}-\sqrt{x+1}$.

(iii) $(f g):[-1,3] \rightarrow R$ is given by $(f g)(x)=f(x) \cdot g(x)=$

$\sqrt{x+1} \cdot \sqrt{9-x^{2}}=\sqrt{(x+1)\left(9-x^{2}\right)}=\sqrt{9+9 x-x^{2}-x^{3}}$

(iv) $\frac{f}{g}:[-1,3] \rightarrow R$ is given by $\left(\frac{f}{g}\right)(\mathrm{x})=\frac{f(x)}{g(x)}=\frac{\sqrt{\mathrm{x}+1}}{\sqrt{9-\mathrm{x}^{2}}}=\sqrt{\frac{x+1}{9-x^{2}}}$.

(v) $\frac{g}{f}:[-1,3] \rightarrow R$ is given by $\left(\frac{g}{f}\right)(\mathrm{x})=\frac{g(x)}{f(x)}=\frac{\sqrt{9}-\mathrm{x}^{2}}{\sqrt{x+1}}=\sqrt{\frac{9-x^{2}}{x+1}}$.

(vi) $2 f-\sqrt{5} g:[-1,3] \rightarrow R$ is given by $(2 f-\sqrt{5} g)(\mathrm{x})=2 \sqrt{x+1}-\sqrt{5}\left(\sqrt{9-x^{2}}\right)$

$=2 \sqrt{x+1}-\sqrt{45-5 x^{2}}$

(vii) $f^{2}+7 f:[-1, \infty] \rightarrow R$ is given by $\left(f^{2}+7 f\right)(\mathrm{x})=f^{2}(x)+7 f(x)$

{Since domain(f) = [-">−- 1, ∞]}

$=(\sqrt{x+1})^{2}+7(\sqrt{x+1})=x+1+7 \sqrt{x+1}$

(viii) $\frac{5}{g}:[-3,3] \rightarrow R$ is defined by $\left(\frac{5}{g}\right)(\mathrm{x})=\frac{5}{\sqrt{9-\mathrm{x}^{2}}}$.

{Since domain(g) = [-">−- 3, 3]}