If f(x)= _e(1-x) and g(x)=[x], then determine the following functions: f 8

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

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

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If $\text{f(x)}=\log_\text{e}(1-\text{x})$ and $\text{g(x)}=[\text{x}],$ then determine the following functions: $\frac{\text{f}}{8}$

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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 $\Big(\frac{\text{f}}{\text{g}}\Big)(\text{x})=\frac{\log_\text{e}(1-\text{x})}{[\text{x}]}$

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