Operations on Functions \(f(x)=\log(1-x)\) and \(g(x)=[x]\)
Question
If
\[ f(x)=\log(1-x) \]and
\[ g(x)=[x] \]then determine each of the following functions:
(i) \(f+g\)
(ii) \(fg\)
(iii) \(\frac{f}{g}\)
(iv) \(\frac{g}{f}\)
Also, find
\((f+g)(-1)\), \((fg)(0)\), \(\left(\frac{f}{g}\right)\left(\frac12\right)\), \(\left(\frac{g}{f}\right)\left(\frac12\right)\)
Solution
Given
\[ f(x)=\log(1-x) \]and
\[ g(x)=[x] \]Here, \([x]\) denotes the greatest integer function.
Domain of \(f(x)\)
For logarithm,
\[ 1-x>0 \] \[ x<1 \]Therefore,
\[ D_f=(-\infty,1) \]Domain of \(g(x)\)
Greatest integer function is defined for all real numbers.
\[ D_g=\mathbb{R} \]Common domain:
\[ (-\infty,1) \](i) \(f+g\)
\[ (f+g)(x)=\log(1-x)+[x] \]Domain:
\[ (-\infty,1) \](ii) \(fg\)
\[ (fg)(x)=[x]\log(1-x) \]Domain:
\[ (-\infty,1) \](iii) \(\frac{f}{g}\)
\[ \left(\frac{f}{g}\right)(x)=\frac{\log(1-x)}{[x]} \]Denominator cannot be zero.
\[ [x]\ne0 \]Since \([x]=0\) for
\[ 0\le x<1 \]these values are excluded.
Domain:
\[ (-\infty,0) \](iv) \(\frac{g}{f}\)
\[ \left(\frac{g}{f}\right)(x)=\frac{[x]}{\log(1-x)} \]Denominator cannot be zero.
\[ \log(1-x)\ne0 \]Since
\[ \log(1-x)=0 \] \[ 1-x=1 \] \[ x=0 \]Therefore, \(x=0\) is excluded.
Domain:
\[ (-\infty,1)\setminus\{0\} \]Function Values
\((f+g)(-1)\)
\[ (f+g)(-1)=\log(1-(-1))+[-1] \] \[ =\log 2-1 \]\((fg)(0)\)
\[ (fg)(0)=[0]\log(1-0) \] \[ =0\cdot \log1 \] \[ =0 \]\(\left(\frac{f}{g}\right)\left(\frac12\right)\)
Since
\[ \left[\frac12\right]=0 \]denominator becomes zero.
Therefore,
\[ \left(\frac{f}{g}\right)\left(\frac12\right) \text{ is not defined} \]