Question
Let f and g be two real functions defined by $\text{f(x)}=\sqrt{\text{x}+1}$ and $\text{g(x)}=\sqrt{9-\text{x}^2}$ Then describe the following functions: f + g

Answer

We have, $\text{f(x)}=\sqrt{\text{x}+1}$ and $\text{g(x)}=\sqrt{9-\text{x}^2}$ We observe that $\text{f(x)}=\sqrt{\text{x}+1}$ is defined for all $\text{x}\geq-1$ So, domain $\text{f}=[-1,\infty]$ Clearly, $\text{g(x)}=\sqrt{9-\text{x}^2}$ is defined for $9-\text{x}^2\geq0$ $\Rightarrow\text{x}^2-9\leq0$ $\Rightarrow\text{x}^2-3^2\leq0$ $\Rightarrow\text{x }\in[-3,3]$ $\therefore\ \text{domain(g)}=[-3,3]$ Now, $\text{domain(f)}\cap\text{domain(g)}$ $=[-1,\infty]\cap[-3,3]$ $=[-1,3]$ f + g : [-1, 3] → R is given by (f + g)(x) = f(x) + g(x) $=\sqrt{\text{x}+1}+\sqrt{9-\text{x}^2}$

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