Find the Domain and Range of \(f(x)=\frac{ax+b}{bx-a}\)

Solution

Domain

Given: $$ f(x)=\frac{ax+b}{bx-a} $$

The denominator cannot be zero.

Therefore, $$ bx-a\ne0 $$

$$ x\ne\frac{a}{b} $$

Hence, the domain is: $$ \mathbb{R}-\left\{\frac{a}{b}\right\} $$

Range

Let $$ y=\frac{ax+b}{bx-a} $$

Cross multiply: $$ y(bx-a)=ax+b $$

$$ bxy-ay=ax+b $$

$$ x(by-a)=ay+b $$

$$ x=\frac{ay+b}{by-a} $$

For \(x\) to exist, $$ by-a\ne0 $$

Therefore, $$ y\ne\frac{a}{b} $$

Hence, the range is: $$ \mathbb{R}-\left\{\frac{a}{b}\right\} $$