Question

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

Answer 1

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]}

Answer 2

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]}