Question

If $f\left(x\right)={\text{log}}_e(1-x)$ and $g\left(x\right)=\left[x\right]$ then find:

(i) $(f+g)\left(x\right)$

(ii) $\left(fg\right)\left(x\right)$

(iii) $\left(\frac{f}{g}\right)\left(x\right)$

(iv) $\left(\frac{g}{f}\right)\left(x\right)$

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

Answer 1

Clearly, ${\text{log}}_e(1-x)$ is defined only when $1-x>0$, i.e., $x<1$

$\therefore \text{dom (f)}=(-\infty ,1)$

Also, $\text{dom (g)}=R$

$\therefore \text{dom (f)}\cap \text{dom (g)}=(-\infty ,1)\cap R=(-\infty ,1)$

(i) $(f+g):(-\infty ,1)\to R$ is given by

$(f+g)\left(x\right)=f\left(x\right)+g\left(x\right)={\text{log}}_e(1-x)+\left[x\right]$

(ii) $\left(fg\right):(-\infty ,1)\to R$ is given by

$\left(fg\right)\left(x\right)=f\left(x\right)\times g\left(x\right)={\text{log}}_e(1-x)\times \left[x\right]$

(iii) $\{x:g(x)=0\}=\{x:[x]=0\}=[0,1)$

$\therefore \text{dom}\left(\frac{f}{g}\right)=\text{dom}\left(f\right)\cap \text{dom}\left(g\right)-\{x:g(x)=0\}$

$=(-\infty ,1)\cap R-[0,1)=(-\infty ,0)$

$\therefore \frac{f}{g}:(-\infty ,0)\to R$ is given by

$\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}=\frac{{\text{log}}_e(1-x)}{\left[x\right]}$

(iv) $\{x:f(x)=0\}=\{x:{\text{log}}_e(1-x)=0\}=\left\{0\right\}$

$\therefore \text{dom}\left(\frac{g}{f}\right)=\text{dom}\left(g\right)\cap \text{dom}\left(f\right)-\{x:f(x)=0\}$

$=R\cap (-\infty ,1)-\left\{0\right\}=(-\infty ,0)\cup (0,1)$

$\therefore \frac{g}{f}:(-\infty ,0)\cup (0,1)\to R$ is given by

$\left(\frac{g}{f}\right)\left(x\right)=\frac{g\left(x\right)}{f\left(x\right)}=\frac{\left[x\right]}{{\text{log}}_e(1-x)}$

Now, we have:

$(f+g)(-1)=f(-1)+g(-1)=[-1]+{\text{log}}_e(1+1)=\left({\text{log}}_{e}2\right)-1$

$\left(fg\right)\left(0\right)=f\left(0\right)\times g\left(0\right)={\text{log}}_e(1-0)\times \left[0\right]=({\text{log}}_{e}1\times 0)=(0\times 0)=0$.

$\left(\frac{f}{g}\right)(-1)=\frac{f(-1)}{g(-1)}=\frac{[-1]}{{\text{log}}_e(1+1)}=\frac{-1}{{\text{log}}_{e}2}$

$\left(\frac{g}{f}\right)\left(\frac{1}{2}\right)=\frac{g\left(\frac{1}{2}\right)}{f\left(\frac{1}{2}\right)}=\frac{\left[\frac{1}{2}\right]}{{\text{log}}_e(1-\frac{1}{2})}=\frac{\left[0.5\right]}{{\text{log}}_e\left(\frac{1}{2}\right)}=0$.