If f(x) = log_e (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), $(\frac{f}{g}) (\frac{1}{2}), (\frac{g}{f}) (\frac{1}{2})$ .

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Solution

Given:
f(x) = log_e (1 − x) and g(x) = [x]
Clearly, f(x) = log_e (1 − x) is defined for all ( 1 $-$ x) > 0.
⇒ 1 > x
⇒ x < 1
⇒ x ∈ ( $-$ ∞, 1)
Thus, domain (f ) = ( $-$ ∞, 1)
Again,
g(x) = [x] is defined for all x ∈ R.
Thus, domain (g) = R
∴ Domain (f) ∩ Domain (g) = ( $-$ ∞, 1) ∩ R
= ( $-$ ∞, 1)

Hence,
(i ) ( f + g ) : ( $-$ ∞, 1) → R is given by ( f + g ) (x) = f (x) + g (x) = log_e (1 − x) + [ x ].

(ii) (fg) : ( $-$ ∞, 1) → R is given by (fg) (x) = f(x).g(x) = log_e (1 − x)[ x ] = [ x ]log_e (1 − x).

(iii) Given:
g(x) = [ x ]
If [ x ] = 0,
x ∈ (0, 1)
Thus,
$domain (\frac{f}{g}) = domain (f) \cap domain (g) - \{x : g (x) = 0\}$
$\frac{f}{g} : (- \infty, 0) \to R is defined by (\frac{f}{g}) (x) = \frac{f (x)}{g (x)} = \frac{\log_{e} (1 - x)}{[x]} .$

(iv) Given:
f(x) = log_e (1 − x)
$\Rightarrow \frac{1}{f (x)} = \frac{1}{\log_{e} (1 - x)}$
$\frac{1}{f (x)}$ is defined if log_e( 1 $-$ x) is defined and log_e(1 – x) ≠ 0.
⇒ (1 $-$ x) > 0 and (1 $-$ x) ≠ 0
⇒ x < 1 and x ≠ 0
⇒ x ∈ ( $-$ ∞, 0)∪ (0, 1)
Thus, $d omain (\frac{g}{f}) = (- \infty, 0) \cup (0, 1)$ = ( $-$ ∞, 1) .
$\frac{g}{f} : (- \infty, 0) \cup (0, 1) \to R defined by (\frac{g}{f}) (x) = \frac{g (x)}{f (x)} = \frac{[x]}{\log_{e} (1 - x)}$

(f + g)( $-$ 1) = f( $-$ 1) + g( $-$ 1)
= log_e{1 – ( $-$ 1)}+ [ $-$ 1]
= log_e 2 – 1
Hence, (f + g)( $-$ 1) = log_e 2 – 1

(fg)(0) = log_e ( 1 – 0) × [0] = 0
$(\frac{f}{g}) (\frac{1}{2}) = does not exist .$
$(\frac{g}{f}) (\frac{1}{2}) = \frac{[\frac{1}{2}]}{\log_{e} (1 - \frac{1}{2})} = 0$

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