What is the domain and range of #f(x) = (x+5)/(x^2+36)#?

1 Answer
Nov 9, 2015

Answer:

The domain is #RR# (all real numbers) and the range is #[[5-sqrt(61))/72,(5+sqrt(61))/72]#
(all real numbers between and including #(5-sqrt(61))/72# and #(5+sqrt(61))/72#).

Explanation:

In the domain, we start with all real numbers, and then remove any which would force us to have the square root of a negative number, or a #0# in the denominator of a fraction.
At a glance, we know that as #x^2 >= 0# for all real numbers, #x^2+36 >= 36 > 0#. Thus the denominator will not be #0# for any real number #x#, meaning the domain includes every real number.

For the range, the easiest way of finding the above values involves some basic calculus. Although it is longer, it is also possible to find them using only algebra, however, with the method detailed below.


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Starting with the function #f(x) = (x+5)/(x^2+36)# we wish to find all possible values of #f(x)#. This is equivalent to finding the domain of the inverse function #f^-1(x)# (a function with the property #f^-1(f(x)) = f(f^-1(x)) = 1#)

Unfortunately, the inverse of #f(x)# in this case is not a function, as it returns 2 values, however, the idea is still the same. We will start with the equation #y = (x+5)/(x^2+36)# and solve for #x# to find the inverse. Next, we will look at the possible values of #y# to find the domain of the inverse, and therefore the range of the original function.

Solving for #x#:
#y = (x+5)/(x^2+36)#
#=> y(x^2 + 36) = x + 5#
#=> yx^2 + 36y = x+ 5#
#=> yx^2 - x + (36y - 5) = 0#

Treating #y# as a constant, we apply the quadratic formula
#ax^2 + bx + c = 0 => x = (-b +- sqrt(b^2 - 4ac))/(2a)#
to obtain

#x = (1 +- sqrt(1 - 4y(36y-5)))/(2y)#

We now need to find the domain of the above expression (note that it is not a function because of the #+-#). Note that by dividing by #y# in the quadratic formula, we lost the possibility of #y=0#, which is clearly possible in the original equation (for #x = -5#). Thus we will disregard the #y# in the denominator of the inverse, and only focus on the square root.

As mentioned previously, we are not allowing for the square root of a value less than 0, and so we have the restriction
#1 - 4y(36y-5) >= 0#
#=> -144y^2 +20y + 1 >= 0#

Using the quadratic formula on #-144y^2 +20y + 1 = 0# we find, after some simplification,
#y = (5+-sqrt(61))/72#

Finally, we can tell that as #|y|# grows large, #-144y^2 +20y + 1# will be less than #0#. Thus we only consider the interval between
#y = (5-sqrt(61))/72# and #y = (5+sqrt(61))/72#

So the allowed values for #y#, and thus the range for #f(x)#, is
#[[5-sqrt(61))/72,(5+sqrt(61))/72]#