Answer:
B. [tex]\left(\dfrac{f}{g}\right)(x)=\dfrac{1}{x-2}[/tex] domain is [tex]x\in \left(-\infty, -\dfrac{5}{3}\right)\cup \left(-\dfrac{5}{3},2\right)\cup (2,\infty).[/tex]
Step-by-step explanation:
If [tex]f(x)=3x+5[/tex] and [tex]g(x)=3x^2-x-10,[/tex] then
[tex]\left(\dfrac{f}{g}\right)(x)=\dfrac{3x+5}{3x^2-x-10}=\dfrac{3x+5}{3x^2+5x-6x-10}=\dfrac{3x+5}{x(3x+5)-2(3x+5)}=\dfrac{3x+5}{(3x+5)(x-2)}.[/tex]
Note that the denominator cannot be equal to 0, so
[tex]3x+5\neq 0 \text{ and } x-2\neq0\\ \\x\neq-\dfrac{5}{3}\text{ and }x\neq 2.[/tex]
If [tex]x\in \left(-\infty, -\dfrac{5}{3}\right)\cup \left(-\dfrac{5}{3},2\right)\cup (2,\infty),[/tex] then
[tex]\left(\dfrac{f}{g}\right)(x)=\dfrac{3x+5}{(3x+5)(x-2)}=\dfrac{1}{x-2}.[/tex]
