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

Sep 23, 2015

Domain:
All real numbers $x$ except $0$ and $3$.
In other words, $x \ne 0$ and $x \ne 3$

Range:
All real numbers.
In other words, $- \infty < f \left(x\right) < + \infty$

#### 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 $f \left(x\right) = \frac{1}{x} + \frac{1}{x - 3}$ 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 $\frac{1}{x}$ becomes undefined, and for $x = 3$, in which case $\frac{1}{x - 3}$ becomes undefined.

Therefore, the domain of this function is:
all real values except $0$ and $3$.
It can be written as
$x \ne 0$ AND $x \ne 3$
Alternatively, it can be written as
$- \infty < x < 0$ OR $0 < x < 3$ OR $3 < x < + \infty$.
One more way:
$\left(- \infty , 0\right) \cup \left(0 , 3\right) \cup \left(3 , + \infty\right)$

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
$f \left(x\right) = a$
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
$\frac{1}{x} + \frac{1}{x - 3} = a$

We assume that $x \ne 0$ and $x \ne 3$ 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 $x \left(x - 3\right)$, we obtain
$x - 3 + x = a x \left(x - 3\right)$
or
$a {x}^{2} + \left(- 3 a - 2\right) x + 3 = 0$

The quadratic equation above has a solution if its discriminant is not negative.
The discriminant of this equation is
${\left(- 3 a - 2\right)}^{2} - 4 \cdot a \cdot 3 = 9 {a}^{2} + 12 a + 4 - 12 a = 9 {a}^{2} + 4$
As we see, the discriminant $9 {a}^{2} + 4$ 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, $y = \frac{1}{x}$ and $y = \frac{1}{x - 3}$. The result would look like
graph{1/x + 1/(x-3) [-10, 10, -5, 5]}