Answer:
The answer is "[tex]\bold{(-\infty, \infty)}[/tex]"
Step-by-step explanation:
Given:
[tex]\to \bold{\frac{\frac{x^2-bx +ax - ab}{x2+bx-ax - ab}}{\frac{x2+bx+ax +ab}{x2-bx-ax+ab}}}[/tex]
[tex]\to \frac{x^2-bx +ax - ab}{x2+bx-ax - ab} \times \frac {x2-bx-ax+ab}{x2+bx+ax +ab}\\\\\to \frac{x(x-b) +a(x - b)}{x(x+b)-a(x +b)} \times \frac {x(x-b)-a(x-b)}{x(x+b)+a(x +b)}\\\\[/tex]
[tex]\to \frac{(x-b) (x+a)}{(x+b)(x-a)} \times \frac {(x-b)(x-a)}{(x+b)(x +a)}\\\\\to \frac{(x-b)}{(x+b)} \times \frac {(x-b)}{(x+b)}\\\\\to \frac{(x-b)^2}{(x+b)^2} \\\\[/tex]
[tex]\to f(x) = \frac{(x-b)^2}{(x+b)^2} \\\\[/tex]
So, the domain are [tex]\bold{(-\infty, \infty)}[/tex]
