Question 3.5.5: Let f and g be real functions , defined by f(x) = √x + 2 and......
Let f and ge real functions, defined by f(x)=\sqrt{x+2} and g(x)=\sqrt{4-x^{2}} .
Find (i)(f+g)(x) \qquad (ii) (f-g)(x)\qquad (iii) (f g)(x)\qquad ( iv) (f f)(x)\\
(v) (g g)(x)\qquad (vi) \left(\frac{f}{g}\right)(x) .
Clearly, f(x)=\sqrt{x+2} is defined for all x \in R such that
\begin{aligned}& x+2 \geq 0, \text { i.e., } x \geq-2 . \\\therefore \quad & \operatorname{dom}(f)=[-2, \infty) .\end{aligned}Again, g(x)=\sqrt{4-x^{2}} is defined for all x \in R such that4-x^{2} \geq 0.
But, 4-x^{2} \geq 0 \Rightarrow x^{2}-4 \leq 0 \Rightarrow(x+2)(x-2) \leq 0 \Rightarrow x \in[-2,2] .
\therefore \quad \operatorname{dom}(g)=[-2,2] .
\therefore \quad \operatorname{dom}(f) \cap \operatorname{dom}(g)=[-2, \infty) \cap[-2,2]=[-2,2] .
(i) (f+g):[-2,2] \rightarrow R is given by
(f+g)(x)=f(x)+g(x)=\sqrt{x+2}+\sqrt{4-x^{2}} .(ii) (f-g):[-2,2] \rightarrow R is given by
(f-g)(x)=f(x)-g(x)=\sqrt{x+2}-\sqrt{4-x^{2}} .(iii) (f g):[-2,2] \rightarrow R is given by
\begin{aligned}(f g)(x) & =f(x) \cdot g(x)=(\sqrt{x+2})\left(\sqrt{4-x^{2}}\right) \\& =\sqrt{(x+2)^{2}(2-x)}=(x+2) \sqrt{(2-x)}.\end{aligned}(iv) (f f):[-2,2] \rightarrow R is given by
(f f)(x)=f(x) \cdot f(x)=(\sqrt{x+2})(\sqrt{x+2})=(x+2) .(v) (g g):[-2,2] \rightarrow R is given by
(g g)(x)=g(x) \cdot g(x)=\left(\sqrt{4-x^{2}}\right)\left(\sqrt{4-x^{2}}\right)=\left(4-x^{2}\right) .(vi) \{x: g(x)=0\}=\left\{x: 4-x^{2}=0\right]=\{x:(2-x)(2+x)=0\}=\{-2,2\} .
\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{\sqrt{x + 2}}{\sqrt{4-x^{2}}}=\frac{\sqrt{2 + x}}{(\sqrt{2 + x})(\sqrt{2-x})}=\frac{1}{(\sqrt{2-x})}.