Answer:
Option B
Step-by-step explanation:
Given that f(x) = 3x+5 and g(x) = 3x^2-x-10
Hence the function f/g = (3x+5)/(3x^2-x-10)
Since denominator cannot be 0 we have
3x^2-x-10 cannot be 0
i.e. x should not take values where 3x^2-x-10 =0
Solving (3x+5)(x-2) =0
x cannot take value as 2 or -5/3
So we can cancel 3x+5 and write
f/g =1/ x-2
Hence x cannot take values as 2
Option B
Option: B is the correct answer.
The function is: [tex](\dfrac{f}{g})(x)=\dfrac{1}{x-2}[/tex]
and the domain is: The set of all real numbers except 2
We are given a function f(x) and g(x) in terms of variable x by:
[tex]f(x)=3x+5[/tex]
and
[tex]g(x)=3x^2-x-10[/tex]
Now, the function
[tex](\dfrac{f}{g})(x)[/tex] is given by:
[tex](\dfrac{f}{g})(x)=\dfrac{f(x)}{g(x)}[/tex]
i.e.
[tex](\dfrac{f}{g})(x)=\dfrac{3x+5}{3x^2-x-10}[/tex]
[tex]3x^2-x-10=3x^2-6x+5x-10\\\\i.e.\\\\3x^2-x-10=3x(x-2)+5(x-2)\\\\i.e.\\\\3x^2-x-10=(3x+5)(x-2)[/tex]
i.e.
[tex](\dfrac{f}{g})(x)=\dfrac{3x+5}{(3x+5)(x-2)}\\\\i.e.\\\\(\dfrac{f}{g})(x)=\dfrac{1}{x-2}[/tex]
Also, the domain of the function is the set of all the real values except where the denominator is equal to zero.
The denominator is equal to zero when x=2.
Hence, the domain is the set of all the real values except 2.
