Answer
$ (f-g)(x)=\frac{x+1}{(x+3)(x-2)}$.
$(-\infty,-3)\cup(-3,-2)\cup(-2,2)\cup(2,\infty)$.
Work Step by Step
The given functions are
$f(x)=\frac{2x-1}{x^2+x-6}$ and $g(x)=\frac{x+2}{x^2+5x+6}$
Use $(f-g)(x)=f(x)-g(x)$.
$\Rightarrow (f-g)(x)=\frac{2x-1}{x^2+x-6}-\frac{x+2}{x^2+5x+6}$.
Factor the denominators of the fractions.
$\Rightarrow x^2+x-6$
Rewrite the middle term $x$ as $3x-2x$.
$\Rightarrow x^2+3x-2x-6$
Group the terms.
$\Rightarrow (x^2+3x)+(-2x-6)$
Factor each group.
$\Rightarrow x(x+3)-2(x+3)$
Factor out $(x+3)$.
$\Rightarrow (x+3)(x-2)$
$\Rightarrow x^2+5x+6$
Rewrite the middle term $5x$ as $3x+2x$.
$\Rightarrow x^2+3x+2x+6$
Group the terms.
$\Rightarrow (x^2+3x)+(2x+6)$
Factor each group.
$\Rightarrow x(x+3)+2(x+3)$
Factor out $(x+3)$.
$\Rightarrow (x+3)(x+2)$
Back substitute the factor into the equation.
$\Rightarrow (f-g)(x)=\frac{2x-1}{(x+3)(x-2)}-\frac{x+2}{(x+3)(x+2)}$.
Cancel common terms.
$\Rightarrow (f-g)(x)=\frac{2x-1}{(x+3)(x-2)}-\frac{1}{(x+3)}$.
The LCD of both the fractions is $(x+3)(x-2)$.
$\Rightarrow (f-g)(x)=\frac{2x-1}{(x+3)(x-2)}-\frac{(x-2)}{(x+3)(x-2)}$.
Add both the numerators because denominators are the same.
$\Rightarrow (f-g)(x)=\frac{2x-1-(x-2)}{(x+3)(x-2)}$.
Simplify.
$\Rightarrow (f-g)(x)=\frac{2x-1-x+2}{(x+3)(x-2)}$.
Add like terms.
$\Rightarrow (f-g)(x)=\frac{x+1}{(x+3)(x-2)}$.
This gives:
$(-\infty,-3)\cup(-3,-2)\cup(-2,2)\cup(2,\infty)$.
