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

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) f+g:(- ,1) defined by (fg)(x)=f(x) (x) = _e(1-x)+[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:
f + g

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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 $\text{x}\in\text{R}$
$\therefore\ \text{Domain(g)}=\text{R}$
$\therefore\ \text{Domain(f)}\cap\text{R}\text{ Domain(g)}=(-\infty,1)\cap\text{R}$
$=(-\infty,1)$
$\text{f}+\text{g}:(-\infty,1)\rightarrow\text{R}$ defined by $(\text{fg)(x)}=\text{f(x)}\times\text{g(x)}$
$=\log_\text{e}(1-\text{x})+[\text{x}]$

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