Question

Let f, g be two real functions defined by $f\left(x\right)=\sqrt{x+1}$ and $g\left(x\right)=\sqrt{9-x^2}$. Then, describe each of the following functions:

(i) $f+g$
(ii) $g-f$
(iii) $fg$
(iv) $\frac{f}{g}$
(v) $\frac{g}{f}$
(vi) $2f-\sqrt{5}g$
(vii) $f^2+7f$
(viii) $\frac{5}{8}$

Answer 1

(i) We have,
$f\left(x\right)=\sqrt{x+1}$ and $g\left(x\right)=\sqrt{9-x^2}$
We observe that $f\left(x\right)=\sqrt{x+1}$ is defined for all $x\ge -1$
So, domain (f) $=[-1,\infty ]$
Clearly, $g\left(x\right)=\sqrt{9-x^2}$ is defined for
$9-x^2\ge 0\Rightarrow x^2-9=\le 0\Rightarrow x^2-3^2\le 0\Rightarrow (x-3)(x+3)\le 0\Rightarrow xϵ[-3,3]$
Now,
$\text{domain}\left(f\right)\cap \text{domain}\left(g\right)$
$=[-1,\infty ]\cap [-3,3]=[-1,3]f+g:[-1,3]\to R\text{is given by (f+g)(x)}=f\left(x\right)+g\left(x\right)=\sqrt{x+1}+\sqrt{9-x^2}$

(ii) We have,
$f\left(x\right)=\sqrt{x+1}$ and $g\left(x\right)=\sqrt{9-x^2}$
We observe that $f\left(x\right)=\sqrt{x+1}$ is defined for all $x\ge -1$
So, domain (f) $=-1,\infty$
Clearly, $g\left(x\right)=\sqrt{9-x^2}$ is defined for
$9-x^2\ge 0\Rightarrow x^2-9\le 0\Rightarrow x^2-3^2\le 0\Rightarrow (x-3)(x+3)\le 0\Rightarrow xϵ[-3,3]$
$\therefore$ domain (g) = [-3, 3]
Now,
$\text{domain}\left(f\right)\cap \text{domain}\left(g\right)=[-1,\infty ]\cap [-3,3]$
= [-1, 3]
$g-f:[-3]\to R$ is given by (g - f)
$\left(x\right)=g\left(x\right)-f\left(x\right)=\sqrt{9-x^2}-\sqrt{x+1}$

(iii) We have,
$f\left(x\right)=\sqrt{x+1}$ and $g\left(x\right)=\sqrt{9-x^2}$
We observethat $f\left(x\right)=\sqrt{x+1}$ is defined for all $x\ge -1$
So, domain (f) $=[-1,\infty )$
Clearly, $g\left(x\right)=\sqrt{9-x^2}$ is defined for
$9-x^2\ge 0\Rightarrow x^2-9\le 0\Rightarrow x^3-3^2\le 0\Rightarrow (x-3)(x+3)\le 0\Rightarrow xϵ[-3,3]$
$\therefore$ domain (g) = [-3, 3]
Now,
$\text{domain}\left(f\right)\cap \text{domain}\left(g\right)=[-1,\infty ]\cap [-3,3]$
= [-1, 3]
$fg:[-,3]\to R$ is given by $\left(fg\right)\left(x\right)=f\left(x\right)\times g\left(x\right)=\sqrt{x+1}\times \sqrt{9-x^2}=\sqrt{9+9x-x^2-x^3}$

(iv) We have,
$f\left(x\right)=\sqrt{x+1}$ and $g\left(x\right)=\sqrt{9-x^2}$
We observe that $f\left(x\right)=\sqrt{x+1}$ is defined for all $x\ge -1$
So, doamin (f) $=[-1,\infty )$
Clearly, $g\left(x\right)=\sqrt{9-x^2}$ is defined for
$9-x^2\ge 0\Rightarrow x^2-9\le 0\Rightarrow x^2-3^2\le 0\Rightarrow (x-3)(x+3)\le 0\Rightarrow xϵ[-3,3]$
$\therefore$ domain (g) = [-3, 3]
Now, $\text{domain}\left(f\right)\cap domain\left(g\right)=[-1,\infty ]\cap [-3,3]$
= [ -1, 3]
We have, $g\left(x\right)=\sqrt{9-x^2}$
$\therefore 9-x^2=0\Rightarrow x^2-9=0\Rightarrow (x-3)(x+3)=0\Rightarrow x=\pm 3$
So, domain $\left(\frac{f}{g}\right)=[-1,3]-[-3,3]=[-1,3]$
$\therefore \frac{f}{g}:[-1,3]\to R$ is given by $\left(\frac{f}{g}\right)\left(x\right)$
$=\frac{f\left(x\right)}{g\left(x\right)}=\frac{\sqrt{x+1}}{\sqrt{9-x^2}}$

(v) We have,
$f\left(x\right)=\sqrt{x+1}$ and $g\left(x\right)=\sqrt{9-x^2}$
We observed that $f\left(x\right)=\sqrt{x+1}$ is defined for all $x\ge -1$
So, domain (f) $=[-1,\infty )$
Clearly, $g\left(x\right)=\sqrt{9-x^2}$ is defined for
$9-x^2\ge 0\Rightarrow x^2-9\le 0\Rightarrow x^2-3^2\le 0\Rightarrow (x-3)(x+3)\le 0\Rightarrow xϵ[-3,3]$
$\therefore$ domain (g) = [-3, 3]
Now,
$\text{domain}\left(f\right)\cap \text{domain}\left(g\right)=[-1,\infty ]\cap [-3,3]$
= [-1, 3]
We have,
$f\left(x\right)=\sqrt{x-1}\therefore \sqrt{x-1}=0\Rightarrow x+1=0\Rightarrow x=-1$
So, domain $\left(\frac{g}{f}\right)=[-1,3]-\{-1\}$
= [-1, 3]
$\therefore \frac{g}{f}:[-1,3]\to R$ is given by $\frac{g}{f}\left(x\right)$
$=\frac{g\left(x\right)}{f\left(x\right)}=\frac{\sqrt{9-x^2}}{\sqrt{x+1}}$

(vi) We have,
$f\left(x\right)=\sqrt{x+1}$ and $g\left(x\right)=\sqrt{9-x^2}$
We observed that $f\left(x\right)=\sqrt{x+1}$ is defined for all $x\ge -1$
So, domain $\left(f\right)=[-1,\infty )$
Clearly, $g\left(x\right)=\sqrt{9-x^2}$ is defined for
$9-x^2\ge 0\Rightarrow x^2-9\le 0\Rightarrow x^2-3^2\le 0\Rightarrow (x-3)(x+3)\le 0\Rightarrow xϵ[-3,3]$
$\therefore$ domain (g) = [-3, 3]
Now,
$\text{domain}\left(f\right)\cap \text{domain}\left(g\right)=[-1,\infty ]\cap [-3,3]$
= [-1, 3]
$2f-\sqrt{5g}:[-,3]\to R$ defined by $(2f-\sqrt{5g})\left(x\right)=2\sqrt{x+1}-\sqrt{5}\sqrt{9-x^2}=2\sqrt{x+1}-\sqrt{45-5x^2}$

(vii) We have,
$f\left(x\right)=\sqrt{x+1}$ and $g\left(x\right)=\sqrt{9-x^2}$
We observed that $f\left(x\right)=\sqrt{x+1}$ is defined for all $x\ge -1$
So, domain $\left(f\right)=[-1,\infty )$
Clearly, $g\left(x\right)=\sqrt{9-x^2}$ is defined for
$9-x^2\ge 0\Rightarrow x^2-9\le 0\Rightarrow x^2-3^2\le 0\Rightarrow (x-3)(x+3)\le 0\Rightarrow xϵ[-3,3]$
$\therefore$ domain (g) = [-3, 3]
Now,
$\text{domain}\left(f\right)\cap \text{domain}\left(g\right)=[-1,\infty ]\cap [-3,3]$
= [-1, 3]
$f^2+7f:[-1,\infty ]\to R$ defined by $(f^2+7f)\left(x\right)=f^2\left(x\right)+7f\left(x\right)$
$[\because D(f\left)\right]=(-1,\infty )$
$=(\sqrt{x+1}{)}^2+7\sqrt{x+1}=x+1+7\sqrt{x+1}$

(viii) We have,
$f\left(x\right)=\sqrt{x+1}$ and $g\left(x\right)=\sqrt{9-x^2}$
We observed that $f\left(x\right)=\sqrt{x+1}$ is defined for all $x\ge -1$
So, domain $\left(f\right)=[-1,\infty )$
Clearly, $g\left(x\right)=\sqrt{9-x^2}$ is defined for
$9-x^2\ge 0\Rightarrow x^2-9\le 0\Rightarrow x^2-3^2\le 0\Rightarrow (x-3)(x+3)\le 0\Rightarrow xϵ[-3,3]$
$\therefore$ domain (g) = [-3, 3]
Now,
$\text{domain}\left(f\right)\cap \text{domain}\left(g\right)=[-1,\infty ]\cap [-3,3]$
= [-1, 3]
$2f-\sqrt{5g}:[-,3]\to R$ defined by $(2f-\sqrt{5g})\left(x\right)=2\sqrt{x+1}-\sqrt{5}\sqrt{9-x^2}=2\sqrt{x+1}-\sqrt{45-5x^2}$

(vii) We have,
$f\left(x\right)=\sqrt{x+1}$ and $g\left(x\right)=\sqrt{9-x^2}$
We observed that $f\left(x\right)=\sqrt{x+1}$ is defined for all $x\ge -1$
So, domain $\left(f\right)=[-1,\infty )$
Clearly, $g\left(x\right)=\sqrt{9-x^2}$ is defined for
$9-x^2\ge 0\Rightarrow x^2-9\le 0\Rightarrow x^2-3^2\le 0\Rightarrow (x-3)(x+3)\le 0\Rightarrow xϵ[-3,3]$
$\therefore$ domain (g) = [-3, 3]
Now,
$\text{domain}\left(f\right)\cap \text{domain}\left(g\right)=[-1,\infty ]\cap [-3,3]$
= [-1, 3]
We have,
$g\left(x\right)=\sqrt{9-x^2}\therefore 9-x^2=0\Rightarrow x^2-9=0\Rightarrow (x-3)(x+3)=0\Rightarrow x=\pm 3$
So, domain $\left(\frac{1}{g}\right)=[-3,3]-\{-3,3\}$
= (-3, 3)
$\therefore \frac{5}{g}=(-3,3)\to R$ defined by $\left(\frac{5}{9}\right)\left(x\right)$
$=\frac{5}{\sqrt{9-x^2}}$