Question

If f(x) = log_e (1 − x) and g(x) = [x], then determine function:

(iv) \[\frac{g}{f}\] Also, find (f + g) (−1), (fg) (0),

\[\left( \frac{f}{g} \right) \left( \frac{1}{2} \right), \left( \frac{g}{f} \right) \left( \frac{1}{2} \right)\]

 

 

Answer 1

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,

(iv) Given:
f(x) = log_e (1 − x)

\[\Rightarrow \frac{1}{f\left( x \right)} = \frac{1}{\log_e \left( 1 - x \right)}\]

\[\frac{1}{f\left( x \right)}\]   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,

\[\text{ domain } \left( \frac{g}{f} \right) = \left( - \infty , 0 \right) \cup \left( 0, 1 \right)\]  = ( - ∞, 1)  .

\[\frac{g}{f}: \left( - \infty , 0 \right) \cup \left( 0, 1 \right) \to \text{ R defined by } \left( \frac{g}{f} \right)\left( x \right) = \frac{g\left( x \right)}{f\left( x \right)} = \frac{\left[ x \right]}{\log_e \left( 1 - x \right)}\]

(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

\[\left( \frac{f}{g} \right)\left( \frac{1}{2} \right) = \text{ does not exist}  . \]

\[\left( \frac{g}{f} \right)\left( \frac{1}{2} \right) = \frac{\left[ \frac{1}{2} \right]}{\log_e \left( 1 - \frac{1}{2} \right)} = 0\]