Question
Find $\text{f}+\text{g},\text{ f}-\text{g},\text{ cf}(\text{c}\in\text{R},\text{c}\neq0),\text{ fg},\frac{1}{\text{f}}$ and $\frac{\text{f}}{\text{g}}$ in the following: $\text{f(x)}=\sqrt{\text{x}-1}\text{ and }\text{g(x)}=\sqrt{\text{x}+1}$

Answer

We have, $\text{f(x)}=\sqrt{\text{x}-1}$ and $\text{g(x)}=\sqrt{\text{x}+1}$ Now, $\text{f}+\text{g}:(1,\infty)\rightarrow\text{R}$ is defined by (f + g)(x) $=\sqrt{\text{x}-1}+\sqrt{\text{x}+1}$ $\text{f}-\text{g}:(1,\infty)\rightarrow\text{R}$ is defined by (f - g)(x) = f(x) - g(x) $=\sqrt{\text{x}-1}-\sqrt{\text{x}+1}$ $\text{cf}:(1,\infty)\rightarrow\text{R}$ is defined by $(\text{cf)(x)}=\text{c}\sqrt{\text{x}-1}$ $(\text{fg}):(1,\infty)\rightarrow\text{R}$ is defined (fg)(x) $=\big(\sqrt{\text{x}-1}\big)\big(\sqrt{\text{x}+1}\big)=\sqrt{\text{x}^2-1}$ $\frac{1}{\text{f}}:(1,\infty)\rightarrow\text{R}$ is defined by $\Big(\frac{1}{\text{f}}\Big)(\text{x})=\frac{1}{\sqrt{\text{x}-1}}$ $\frac{\text{f}}{\text{g}}:\big(1,\infty)\rightarrow\text{R}$ is defined by $\Big(\frac{\text{f}}{\text{g}}\Big)(\text{x})=\sqrt{\frac{\text{x}+1}{\text{x}-1}}$

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