Question
If $\text{f(x)}=\log_\text{e}(1-\text{x})$ and $\text{g(x)}=[\text{x}],$ then determine the following functions:
$\frac{\text{f}}{8}$
$\frac{\text{f}}{8}$
✓
Answer
We have,
$\text{f(x)}=\log_\text{e}(1-\text{x})$
and $\text{g(x)}=[\text{x}]$
$\text{f(x)}=\log_\text{e}(1-\text{x})$ is defined, if 1 - x > 0
$\Rightarrow1>\text{x}$
$\Rightarrow\text{x}<1$
$\Rightarrow\text{x}\in(-\infty,1)$
$\therefore\text{ Domain(f)}=(-\infty,1)$
$\text{g(x)}=[\text{x}]$ is defined for all $\text{x}\in\text{R}$
$\therefore\ \text{Domain(g)}=\text{R}$
$\therefore\ \text{Domain(f)}\cap\text{R}\text{ Domain(g)}=(-\infty,1)\cap\text{R}$
$=(-\infty,1)$
$\text{g(x)}=[\text{x}]$
$\therefore\ [\text{x}]=0$
$\Rightarrow\text{x}\in(0,1)$
So, domain $\Big(\frac{\text{f}}{\text{g}}\Big)=\text{domain(f)}\cap\text{domain(g)}-\{\text{x}:\text{g(x)}=0\}$
$\therefore\ \frac{\text{f}}{\text{g}}:(-\infty,0)\rightarrow\text{R}$ defined by $\Big(\frac{\text{f}}{\text{g}}\Big)(\text{x})=\frac{\log_\text{e}(1-\text{x})}{[\text{x}]}$
$\text{f(x)}=\log_\text{e}(1-\text{x})$
and $\text{g(x)}=[\text{x}]$
$\text{f(x)}=\log_\text{e}(1-\text{x})$ is defined, if 1 - x > 0
$\Rightarrow1>\text{x}$
$\Rightarrow\text{x}<1$
$\Rightarrow\text{x}\in(-\infty,1)$
$\therefore\text{ Domain(f)}=(-\infty,1)$
$\text{g(x)}=[\text{x}]$ is defined for all $\text{x}\in\text{R}$
$\therefore\ \text{Domain(g)}=\text{R}$
$\therefore\ \text{Domain(f)}\cap\text{R}\text{ Domain(g)}=(-\infty,1)\cap\text{R}$
$=(-\infty,1)$
$\text{g(x)}=[\text{x}]$
$\therefore\ [\text{x}]=0$
$\Rightarrow\text{x}\in(0,1)$
So, domain $\Big(\frac{\text{f}}{\text{g}}\Big)=\text{domain(f)}\cap\text{domain(g)}-\{\text{x}:\text{g(x)}=0\}$
$\therefore\ \frac{\text{f}}{\text{g}}:(-\infty,0)\rightarrow\text{R}$ defined by $\Big(\frac{\text{f}}{\text{g}}\Big)(\text{x})=\frac{\log_\text{e}(1-\text{x})}{[\text{x}]}$
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