You are correct! The problems that can arise with a function domain are:
- Denominators that become zero
- Even-degree roots with negative input
- Logarithms with negative or zero input
In this case, you have a denominator, and you don't have roots nor logarithms. This means that your only concern must be the denominator, specifically, it cannot be zero.
And you simply have

So, the domain of this function includes every number except 3.
Here is an example I did of the problem hope it helps let me know if you need anything else~!
~<span>Cecildoesscience </span>
Expand the equation.
Multiply (distribute) the first numbers of each set, outer numbers of each set, inner numbers of each set, and the last numbers of each set.
Combine like terms.
Solve the equation and simplify, if needed.
Answer:
d. 15
Step-by-step explanation:
Putting the values in the shift 2 function
X1 + X2 ≥ 15
where x1= 13, and x2=2
13+12≥ 15
15≥ 15
At least 15 workers must be assigned to the shift 2.
The LP model questions require that the constraints are satisfied.
The constraint for the shift 2 is that the number of workers must be equal or greater than 15
This can be solved using other constraint functions e.g
Putting X4= 0 in
X1 + X4 ≥ 12
gives
X1 ≥ 12
Now Putting the value X1 ≥ 12 in shift 2 constraint
X1 + X2 ≥ 15
12+ 2≥ 15
14 ≥ 15
this does not satisfy the condition so this is wrong.
Now from
X2 + X3 ≥ 16
Putting X3= 14
X2 + 14 ≥ 16
gives
X2 ≥ 2
Putting these in the shift 2
X1 + X2 ≥ 15
13+2 ≥ 15
15 ≥ 15
Which gives the same result as above.