Question

If $f\left(x\right)=lo{g}_e(1-x)$ and $g\left(x\right)=[x$, then determine each of the following functions :

(i) $f+g$
(ii) $fg$
(iii) $\frac{f}{g}$
(iv) $\frac{g}{f}$
Also, find $(f+g)(-1),\left(fg\right)\left(0\right),\left(\frac{f}{g}\right)\left(\frac{1}{2}\right),\left(\frac{g}{f}\right)\left(\frac{1}{2}\right)$

Answer 1

We have
$f\left(x\right)=lo{g}_e(1-x)$ and $g\left(x\right)=\left[x\right]$
$f\left(x\right)=lo{g}_e(1-x)$ is defined, if $1-x>0$
$\Rightarrow 1>x\Rightarrow x<1\Rightarrow xϵ(-\infty ,1)$
g(x) = [x] is defined for all $xϵR$
$\therefore$ Domain (g) = R
$\therefore \text{Domain}\left(f\right)\cap R\text{Domain}\left(g\right)=(-\infty ,1)\cap R=(-\infty ,1)$

(i) $f+g:(-\infty ,1)\to R$ defined by
$(f+g)\left(x\right)=f\left(x\right)+g\left(x\right)=lo{g}_e(1-x)+\left[x\right]$

(ii) $fg:(-\infty ,1)\to R$ defined by $\left(fg\right)\left(x\right)=f\left(x\right)\times g\left(x\right)$
$=lo{g}_e(1-x)\times \left[x\right]=\left[x\right]lo{g}_e(1-x)$

(iii) g(x) = [x]
$\therefore$ [x] = 0
$\Rightarrow xϵ(0,1)$
So, domain $\left(\frac{f}{g}\right)=\text{domain}\left(f\right)\cap \text{domain}\left(g\right)-\{x:g(x)=0\}=(-\infty ,0)$
$\therefore \frac{f}{g}:(-\infty ,0)\to$ defined by $\left(\frac{f}{g}\right)\left(x\right)=\frac{lo{g}_e(1-x)}{\left[x\right]}$

(iv) We have,
$f\left(x\right)=lo{g}_e(1-x)$
$\Rightarrow \frac{1}{f\left(x\right)}=\frac{1}{lo{g}_e(1-x)}$ is defined, and $lo{g}_e(1-x)\ne 0$
$\Rightarrow 1-x>0$ and $1-x\ne 0$
$\Rightarrow x<1$ and $x\ne 0$
$\Rightarrow xϵ(-\infty ,0)\cup (0,1)$
$\therefore$ Domain $\left(\frac{g}{f}\right)=(-\infty ,0)\cup (0,1)$
$\frac{g}{f}:(-\infty ,0)\cup (0,1)\to R$ defined by
$\left(\frac{g}{f}\right)\left(x\right)=\frac{\left[x\right]}{lo{g}_e(1-x)}$
Now,
$(f+g)(-1)=f(-1)+g(-1)=lo{g}_e(1-(-1\left)\right)+[-1]=lo{g}_{e}2-1\Rightarrow (f+g)(-1)=lo{g}_{e}2-1$

(v) $f_g\left(0\right)=lo{g}_e(1-0)\times \left[0\right]$

(vi) $\left(\frac{f}{g}\right)\left(\frac{1}{2}\right)=$ does not exist

(vii) $\left(\frac{g}{f}\right)\left(\frac{1}{2}\right)=\frac{\left(\frac{1}{2}\right)}{lo{g}_e(1-\frac{1}{2})}=0$