How do you find the domain and range of #g(x) = x/(x^2 - 16)#?

2 Answers
Mar 17, 2018

Answer:

The domain of #g(x)# is # x in RR-{-4,4}#.
The range is #g(x) in RR#

Explanation:

As you cannot divide by #0#, the denominator is #!=0#

Therefore,

#x^2-16!=0#, #=># #x!=-4# and #x!=4#

The domain of #g(x)# is #x in RR-{-4,4}#

To calculate the range, proceed as follows

Let #y=x/(x^2-16)#

#y(x^2-16)=x#

#yx^2-x-16y=0#

This is a quadratic equation in #x#, and in order to have solutions,
the discriminant #>=0#

#a=y#

#b=-1#

#c=-16y#

#Delta=b^2-4ac=(-1)^2-4(y)(-16y)=1+64y^2#

#AA y in RR#, #=>#, #Delta>=0#

Therefore,

The range is #g(x) in RR#

graph{x/(x^2-16) [-10, 10, -5, 5]}

Mar 17, 2018

Answer:

Domain: #(-oo, -4)uu(-4, 4)uu(4, oo)#
Range: #(-oo, oo)#

Explanation:

Given: #x/(x^2 - 16)#

First factor the denominator since #(x^2-16)# is the difference of squares:

#x/(x^2 - 16) = x/((x-4)(x+4))#

Find the Domain - valid input - usually #x#
For most functions, the domain is #(-oo, oo)#, the set of all reals. There are a number of factors that can cause this domain to be limited. Here are a few possibilities:

  • a radical such as a square root - limits the domain
  • a denominator - can produce holes and/or vertical asymptotes
  • inverse trigonometry functions
  • natural log function #(y = ln x)#

In your example, the vertical asymptotes are the cause. When the denominator function #D(x) = 0#, the vertical asymptotes are found to be at #x = +-4#

Domain: #(-oo, -4)uu(-4, 4)uu(4, oo)#

Find the Range - valid output - usually #y#
For most functions, the range is also #(-oo, oo)#, the set of all reals. There are a number of factors that can cause this range to be limited. Here are a few possibilities:

  • a radical such as a square root - limits the range
  • a quadratic or even powered function can limit the range. The vertex will be a minimum or a maximum
  • absolute value functions can have a vertex
  • a rational function (has a numerator and denominator) can have a horizontal asymptote
  • a natural exponential function (#y = e^x#)

In your example, w have a rational function. The degree of the numerator function = 1 #(n = 1)#and the degree of the denominator function = 2, #(m = 2)#. When #n < m# there is a horizontal asymptote at #y = 0#.

Range: #(-oo, 0) uu (0, oo)#

But you can see from the graph below that the point #(0,0)# exists. This means the domain is actually #(-oo, oo)#

How would you know this without graphing the function? Create a table of values.
#ul(x|-8" "|-4" "|-2" "|0" "|2" "|4" "|8 " ")#
#y|-1/6" "|un" "|1/6" "|0" "|-1/6""|un" "|1/6#

#un# = undefined

graph{x/(x^2-16) [-10, 10, -5, 5]}