Domain and range of rational functions

Recall that a rational function is a function of the form $y=\frac{P\left(x\right)}{Q\left(x\right)}$y=P(x)Q(x)​ where $P\left(x\right)$P(x) and $Q\left(x\right)$Q(x) are polynomials and $Q\left(x\right)\ne0$Q(x)≠0.

 

The Domain 

Setting the denominator function to zero

When dealing with the domain of a rational function, we need to locate any points of discontinuity. As these are points where the function is not defined, we simply solve $Q\left(x\right)=0$Q(x)=0 to find them. If there are no solutions to this equation (and it frequently happens), then there are no points of discontinuity and the curve is said to be continuous across its entire domain.

 

A cautionary note

In most cases if there is a point of discontinuity, the curve will exhibit asymptotic behaviour on either side of it.

As a word of caution however, asymptotic behaviour is not always guaranteed.  This is because some of the solutions to $Q\left(x\right)=0$Q(x)=0 may also be solutions to $P\left(x\right)=0$P(x)=0.

Suppose for example we find a value of $x$x, say $x=h$x=h, where both $Q\left(h\right)=0$Q(h)=0 and $P\left(h\right)=0$P(h)=0. In that case, the curve will, at first glance, look to smoothly cross the line $x=h$x=h without any asymptotic behaviour at all. However, what actually exists is a "hole" in the curve at $x=h$x=h. See example 2 below for a demonstration of this.

 

Domain Example 1  

Find the domain of the function $y=\frac{x^2}{x^2-1}$y=x2x2−1​. 

Here, $Q\left(x\right)=x^2-1$Q(x)=x2−1 and so we simply solve the equation $x^2-1=0$x2−1=0. The solutions are $x=\pm1$x=±1, so there are two discontinuities to deal with. Neither $x=1$x=1, nor $x=-1$x=−1 are solutions to $P\left(x\right)=0$P(x)=0, so the curve will exhibit asymptotic behaviour around these points.

The domain is thus given formally as $\left\{x:x\in\Re\setminus x=\pm1\right\}${x:x∈ℜ∖x=±1}. That is to say the domain includes all real numbers except $x=1$x=1 and $x=-1$x=−1. 

Here is the curve showing these discontinuities. Note that domain considerations are only one part of an overall strategy to sketch these functions. Calculus techniques are often needed to ascertain many other features.

 

domain Example 2

Find the domain of the function $y=\frac{x-1}{x^2-x}$y=x−1x2−x​

Since $Q\left(x\right)=x^2-x$Q(x)=x2−x, we need to solve $x^2-x=0$x2−x=0 so that $x\left(x-1\right)=0$x(x−1)=0 and thus $x=0,1$x=0,1. Note here that the common factor $\left(x-1\right)$(x−1) exists between $P\left(x\right)$P(x) and $Q\left(x\right)$Q(x), and so the curve will behave in a different way near $x=0$x=0 than at $x=1$x=1. 

The domain is thus given formally as $\left\{x:x\in\Re\setminus x=0,1\right\}${x:x∈ℜ∖x=0,1}.

Here it is:

This example is instructive in the sense that, apart from the discontinuity at $x=1$x=1, the curve is exactly the same as that given by $y=\frac{1}{x-1}$y=1x−1​.

You may be wondering why, in the definition of a rational function, these common factors are not simply divided out. The simple answer is that, as a modelling tool, it is convenient to be able to, from time to time, exclude certain values of $x$x from the domain. So common factors in $P\left(x\right)$P(x) and $Q\left(x\right)$Q(x) are left in the definition.

 

domain Example 3   

Find the domain of the function given by $y=\frac{12}{x^2+4x-21}$y=12x2+4x−21​.

Without concerning ourselves with a sketch, the exclusions in the domain are found by solving the quadratic equation $x^2+4x-21=0$x2+4x−21=0. 

$x^2+4x-21$x2+4x−21	$=$=	$0$0
$\left(x-3\right)\left(x+7\right)$(x−3)(x+7)	$=$=	$0$0
$\therefore$∴     $x$x	$=$=	$3,-7$3,−7

Hence the domain is given by $\left\{x:x\in\Re\setminus x=3,-7\right\}${x:x∈ℜ∖x=3,−7}.

 

The Range 

In general terms it is very difficult to establish the range of a rational function without access to calculus tools. However, for some functions, the range can be established in two ways:

• By considering its extreme behaviour (based on horizontal asymptotes)
• By rearrangement of the function so that $x$x becomes the subject of the formula 
• By utilising computer graphing software to pinpoint local minimum and maximum turning points.  

Let's examine a few examples

Range example 1

Find the range of the function given by $y=\frac{12}{x-5}$y=12x−5​.

The variable fraction $\frac{12}{x-5}$12x−5​ can never have a value $0$0, because of the constant numerator $12$12. There is no other exclusion from the range possible and this can be seen by rearrangement:

$y$y	$=$=	$\frac{12}{x-5}$12x−5​
$x-5$x−5	$=$=	$\frac{12}{y}$12y​
$x$x	$=$=	$5+\frac{12}{y}$5+12y​

Apart from $y=0$y=0, any value of $y$y will produce a value of $x$x, and so the only exclusion is $y=0$y=0. In fact, $y=0$y=0 is the horizontal asymptote. 

 

range Example 2

Find the range of the function given by $y=\frac{12}{x^2-1}$y=12x2−1​.

Again we can see that $y\ne0$y≠0 is excluded from the range, but we need to carefully explore any other exclusions. As a strategy, we will rearrange the equation to express $x$x as the subject of the formula:

$y$y	$=$=	$\frac{12}{x^2-1}$12x2−1​
$x^2-1$x2−1	$=$=	$\frac{12}{y}$12y​
$x^2$x2	$=$=	$1+\frac{12}{y}$1+12y​
$x$x	$=$=	$\pm\sqrt{1+\frac{12}{y}}$±√1+12y​

Thus we need to ensure that $1+\frac{12}{y}\ge0$1+12y​≥0 and so solving this carefully we have:

$1+\frac{12}{y}$1+12y​	$\ge$≥	$0$0
$\frac{12}{y}$12y​	$\ge$≥	$-1$−1

Clearly, if $y>0$y>0, the left hand side is positive, and all positive numbers are greater than $-1$−1. This means that part of our solution is given by $y>0$y>0.

In the case where $y<0$y<0. and being mindful of the need to switch the inequality sign when multiplying by a negative quantity, we have:

$12$12	$\le$≤	$-y$−y
$\therefore$∴     $y$y	$\le$≤	$-12$−12

Hence the range is given as  $\left\{y:y\le-12,y>0,y\in\Re\right\}${y:y≤−12,y>0,y∈ℜ}.

The graph is shown below. The exclusion zone for the range is shaded in, so our algebra accords with the graph. The local maximum stationary point is given as $\left(12,0\right)$(12,0), and, given a graph of this function, by observation we can immediately write down the range.

However, for many rational functions, these local extrema are not so clear cut, and thus the only way to proceed is either with an algebraic approach (as we have done here) or else resort to calculus techniques.

 

range example 3 (extension) 

Find the range of $y=\frac{2x}{x^2+4}$y=2xx2+4​

Before we think about the range, note that for all $x$x, $Q\left(x\right)\ne0$Q(x)≠0 and the domain includes all reals and the function is continuous everywhere.

When $x=0$x=0, $y=0$y=0 and the value of the function at $x=a$x=a given by $\frac{2a}{a^2+4}$2aa2+4​ is the negative of the value of the function at $x=-a$x=−a, namely $\frac{-2a}{a^2+4}$−2aa2+4​. The means that the function is odd and exhibits rotational symmetry about the origin.  

In terms of the range, the strategy of rearranging the equation to make $x$x the subject becomes a more difficult exercise. Never-the-less it is possible to do, by considering the equation as a quadratic in $x$x. 

$y$y	$=$=	$\frac{2x}{x^2+4}$2xx2+4​
$yx^2+4y$yx2+4y	$=$=	$2x$2x
$yx^2-2x+4y$yx2−2x+4y	$=$=	$0$0
$\therefore$∴     $x$x	$=$=	$\frac{2\pm\sqrt{4-16y^2}}{2y}$2±√4−16y22y​
	$=$=	$\frac{1\pm\sqrt{1-4y^2}}{y}$1±√1−4y2y​

Hence the range is determined by solving the inequality $1-4y^2\ge0$1−4y2≥0. 

The inequality can be expressed as $\left(1-2y\right)\left(1+2y\right)\ge0$(1−2y)(1+2y)≥0 and with a little thought, the solution can be quite easily be deduced as  $-\frac{1}{2}\le y\le\frac{1}{2}$−12​≤y≤12​.  

Hence the range is given by $\left\{y:-\frac{1}{2}\le y\le\frac{1}{2},y\in\Re\right\}${y:−12​≤y≤12​,y∈ℜ}.

Of course with modern software, we might determine that the range was restricted to values within this interval. However, even with the graph, an algebraic approach verifies that the bounds are indeed given by $\pm\frac{1}{2}$±12​ (and not values that perhaps may be slightly different).

Worked Examples

Question 1

Question 2

QUESTION 3