Question

How do you find the domain and range of `f(x)= 1/x + 5/(x-3)`?

Answer 1

Domain:
All real numbers `x` except `0` and `3`.
In other words, `x!=0` and `x!=3`

Range:
All real numbers.
In other words, `-oo < f(x) < +oo`

Explanation:

It is traditionally assumed that functions like this, unless specifically mentioned otherwise, are defined for real numbers as argument, having values also among real numbers.

Domain of a real function is a set of values where this function is defined.
The function `f(x) = 1/x + 1/(x-3)` is defined for all real values of argument `x` except those where the denominator of one of its two terms equals to `0`.
This happens only for `x=0`, in which case `1/x` becomes undefined, and for `x=3`, in which case `1/(x-3)` becomes undefined.

Therefore, the domain of this function is:
all real values except `0` and `3`.
It can be written as
`x != 0` AND `x != 3`
Alternatively, it can be written as
`-oo < x < 0` OR `0 < x < 3` OR `3 < x < +oo`.
One more way:
`(-oo,0) uu (0,3) uu (3,+oo)`

Range of a real function is a set of values that this function can take while its argument takes all the values from the domain.
To determine the range, let's try to resolve an equation
`f(x) = a`
for any value `a`. Every value for which this equation has a solution (a value of `x` from the domain) belongs to a range.

So, let's try to find all `a`, for which there is a solution of the equation
`1/x + 1/(x-3) = a`

We assume that `x!=0` and `x!=3` since we have already excluded these values of argument, they cannot be solutions, even if we come to these values for some `a`.

Multiplying the equation by `x(x-3)`, we obtain
`x-3 + x = ax(x-3)`
or
`ax^2+(-3a-2)x+3 = 0`

The quadratic equation above has a solution if its discriminant is not negative.
The discriminant of this equation is
`(-3a-2)^2 - 4*a*3 = 9a^2+12a+4-12a = 9a^2+4`
As we see, the discriminant `9a^2+4` is always positive for any real `a`. Therefore, any real value can be a value of our function. In other words, the range of our function is all real values.

An interesting exercise would be to graph this function. I suggest to add two graphs, `y=1/x` and `y=1/(x-3)`. The result would look like
graph{1/x + 1/(x-3) [-10, 10, -5, 5]}