If f(x)= _e(1-x) and g(x)=[x], then determine the following functions: g f

We have, f(x)= _e(1-x) and g(x)=[x] f(x)= _e(1-x) is defined, if 1 - x > 0 1>x 0 and 1-x 0 <1 and x 0 (- ,0) (0,1) domain ( g f )=(- ,0) (0,1) g f :(- ,0) (0,1) defined by ( g f )(x)= [x] _e(1-x)

If f(x)= _e(1-x) and g(x)=[x], then determine the following funct…

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Question

If $\text{f(x)}=\log_\text{e}(1-\text{x})$ and $\text{g(x)}=[\text{x}],$ then determine the following functions: $\frac{\text{g}}{\text{f}}$

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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 $\text{x}\in\text{R}$ $\therefore\ \text{Domain(g)}=\text{R}$ $\therefore\ \text{Domain(f)}\cap\text{R}\text{ Domain(g)}=(-\infty,1)\cap\text{R}$ $=(-\infty,1)$ we have, $\text{f(x)}=\log_\text{e}(1-\text{x})$ $\Rightarrow\frac{1}{\text{f(x)}}=\frac{1}{\log_\text{e}(1-\text{x})}$ $\therefore\ \frac{1}{\text{f(x)}}$ is defined if $\log_\text{e}(1-\text{x})$ is defined and $\log_\text{e}(1-\text{x})\neq0$ $\Rightarrow1-\text{x}>0$ and $1-\text{x}\neq0$ $\Rightarrow\text{x}<1$ and $\text{x}\neq0$ $\Rightarrow\text{x}\in(-\infty,0)\cap(0,1)$ $\therefore\ \text{domain}\Big(\frac{\text{g}}{\text{f}}\Big)=(-\infty,0)\cup(0,1)$ $\frac{\text{g}}{\text{f}}:(-\infty,0)\cap(0,1)\rightarrow\text{R}$ defined by $\Big(\frac{\text{g}}{\text{f}}\Big)(\text{x})=\frac{[\text{x}]}{\log_\text{e}(1-\text{x})}$