Question

Let f and 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) f g
(iv) $\frac{f}{g}$
(v) $\frac{g}{f}$
(vi) $2f-\sqrt{5}g$
(vii) f² + 7f
(viii) $\frac{5}{8}$

Answer 1

Given:
$f\left(x\right)=\sqrt{x+1}\mathrm{and}g\left(x\right)=\sqrt{9-x^2}$
Clearly, $f\left(x\right)=\sqrt{x+1}$ is defined for all x ≥ $-$1.
Thus, domain (f) = [ $-$1, ∞]
Again,

$g\left(x\right)=\sqrt{9-x^2}$ is defined for
9 $-$x² ≥ 0 ⇒ x² $-$ 9 ≤ 0
⇒ x² $-$ 3² ≤ 0
⇒ (x + 3)(x $-$ 3) ≤ 0
⇒ $x\in \left[-3,3\right]$
Thus, domain (g) = [$-$ 3, 3]
Now,
domain ( f ) ∩ domain( g ) = [ $-$1, ∞] ∩ [$-$ 3, 3]
= [ $-$1, 3]
(i) ( f + g ) : [ $-$ 1 , 3] → R is given by ( f + g ) (x) = f (x) + g (x) = $\sqrt{x+1}+\sqrt{9-x^2}$ .

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

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

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

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

(vi) $2f-\sqrt{5}g:\left[-1,3\right]\to R\mathrm{is}\mathrm{given}\mathrm{by}\left(2f-\sqrt{5}g\right)\left(\mathrm{x}\right)=2\sqrt{x+1}-\sqrt{5}\left(\sqrt{9-x^2}\right)$
$=2\sqrt{x+1}-\sqrt{45-5x^2}$ .

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

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