Question

Find the domain and range of the real valued function:

(i) \[f\left( x \right) = \frac{ax + b}{bx - a}\]

 

Answer 1

(i)
Given: 

\[f\left( x \right) = \frac{ax + b}{bx - a}\]

Domain of f : Clearly,  f (x) is a rational function of x as

\[\frac{ax + b}{bx - a}\] is a rational expression.

Clearly, f (x) assumes real values for all x except for all those values of x for which ( bx-a) = 0, i.e. bx = a. 

\[\Rightarrow x = \frac{a}{b}\] 

Hence, domain ( f ) =\[R - \left\{ \frac{a}{b} \right\}\]

Range of f :

Let f (x) = y ⇒ (ax + b) = y (bx -a)
⇒ (ax + b) = (bxy -ay)
⇒ b + ay = bxy -ax
⇒ b + ay = x(by - a) 

 

\[\Rightarrow x = \frac{b + ay}{by - a}\] 

Clearly, f (x) assumes real values for all x except for all those values of x for which ( by - a) = 0, i.e. by = a.

\[\Rightarrow y = \frac{a}{b}\] 

Hence, range ( f ) =\[R - \left\{ \frac{a}{b} \right\}\]