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

1 Answer
Sep 23, 2015

Answer:

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]}