Question 3.5.7: If f(x) = loge(1 - x) and g(x) = [x] then find: (i) (f + g)(......
If f(x)=\log _{e}(1-x) and g(x)=[x] then find:
(i) (f+g)(x)\qquad (ii) (f g)(x)\qquad (iii) \left(\frac{f}{g}\right)(x)\qquad (iv) \left(\frac{g}{f}\right)(x) .
Also find (f+g)(-1),(f g)(0),\left(\frac{f}{g}\right)(-1),\left(\frac{g}{f}\right)\left(\frac{1}{2}\right) .
Clearly, \log _{e}(1-x) is defined only when 1-x>0 , i.e., x<1 .
\therefore \quad \operatorname{dom}(f)=(-\infty, 1) \text {. }Also, \operatorname{dom}(g)=R .
\therefore \quad \operatorname{dom}(f) \cap \operatorname{dom}(g)=(-\infty, 1) \cap R=(-\infty, 1) .
(i) (f+g)(-\infty, 1) \rightarrow R is given by
(f+g)(x)=f(x)+g(x)=\log _{e}(1-x)+[x](ii) (f g):(-\infty, 1) \rightarrow R is given by
(f g)(x)=f(x) \times g(x)=\left\{\log _{e}(1-x)\right\} \times[x] .(iii) \{x: g(x)=0\}=\{x:[x]=0\}=[0,1) .
\therefore \quad \frac{f}{g}:(-\infty, 0) \rightarrow R is given by
\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{\log _{e}(1-x)}{[x]} .(iv) \{x: f(x)=0\}=\left\{x: \log _{e}(1-x)=0\right\}=\{0\} .
Now, we have: