4.03 Domain and range

Remember that functions and relations describe ways of linking two sets of data. In addition, a relation between $x$x and $y$y is a function if there is only one $y$y value associated with each $x$x value.

For example, $\left\{\left(1,2\right),\left(5,3\right),\left(2,-7\right),\left(5,-1\right)\right\}${(1,2),(5,3),(2,−7),(5,−1)} is not a function as the same $x$x value of $5$5 can be related to the $y$y values of both $3$3 and $-1$−1. Recall that graphically, the vertical line test can be used to see if a relation is a function.

Sometimes, a relation can be restricted over a certain interval to then define a function. To do this, the domain and range of the relation need to be considered. 

Domain and range provide a way to describe the set of inputs and outputs for relations. 

There are a number of ways to find the domain and range of a relation. One is to look at the coordinates given and list the possible values. For example, in the previous example, the domain is $\left\{1,5,2\right\}${1,5,2} and the range is $\left\{2,3,-7,-1\right\}${2,3,−7,−1}. Notice how repeated values are not included and order is not important, as we only care about the possible values of $x$x and $y$y. 

The other method is to look at a relation graphically, and see how 'wide' or 'long' the graph is:

Horizontally this graph spans from $-1$−1 to $1$1, so this means that the domain can be expressed as $-1\le x\le1$−1≤x≤1. Similarly, the graph spans vertically from $-2$−2 to $2$2 so the range may be written as $-2\le y\le2$−2≤y≤2.

 

Practice questions

Question 1

question 2

 

Implied domain of an equation

A relation or function may be given as an equation, for example $f(x)=x^2+1$f(x)=x2+1. The equation may have implied restrictions because of the type of equation, or even restrictions implied or defined by the context of the problem.

When looking for the implied domain (sometimes called the maximal domain) of an equation, ask "Is there any input value where the equation will be undefined?". Examples of this are square roots of negative numbers or denominators equal to zero.

Exploration

The function $f\left(x\right)=\sqrt{x}$f(x)=√x, defined for real numbers, is undefined for negative real numbers. Hence, its domain is: $x\ge0$x≥0.

The function $f\left(x\right)=\frac{1}{x}$f(x)=1x​ cannot be evaluated at zero. Hence, its domain is the compound interval: $\left(-\infty,0\right)\cup\left(0,\infty\right)$(−∞,0)∪(0,∞)

The first example, $f\left(x\right)=x^2+1$f(x)=x2+1 has no exclusions and so its implied domain includes all real numbers: $\left(-\infty,\infty\right)$(−∞,∞).

Secondly, there might be what could be called user-defined restrictions where certain restrictions are put onto the domain so that the function makes sense in the real world.

For example, consider a function $f\left(x\right)=3.5x$f(x)=3.5x used to determine the revenue on selling $x$x apples. Here, the domain is given by the set of positive integers because the number of apples can't be negative. Consider another function $f\left(t\right)=200t-4.9t^2$f(t)=200t−4.9t2 where $t$t is the elapsed time in seconds after the stroke of midnight on 31 December 2015. Negative time makes no sense, so restrict the input to the time domain  $t\ge0$t≥0. 

 

Worked example

Find the implied domain of $f\left(x\right)=\sqrt{x-x^2}$f(x)=√x−x2, defined over the real numbers. 

If we try $x=2$x=2, we immediately find a problem, since $f\left(2\right)=\sqrt{2-2^2}=\sqrt{-2}$f(2)=√2−22=√−2. Clearly the implied domain is some subset of the real numbers that doesn't include $2$2. But how do we find it?

We know that we need $x-x^2\ge0$x−x2≥0, or after factorising, $x\left(1-x\right)\ge0$x(1−x)≥0. This means that the product of the two numbers $x$x and $1-x$1−x must be positive. So, $x$x must be confined to the interval $0\le x\le1$0≤x≤1.

Any number outside that interval will cause either $x$x or $1-x$1−x (but never both) to become negative, and hence the product will become negative as well. So the implied domain is given by $0\le x\le1$0≤x≤1.

 

Range 

To find the range, try one of the following strategies:

• Graph the equation using technology and observe the range.
• Use knowledge of the type of graph that the equation produces to understand if there are any restrictions in the vertical direction. E.g. Find turning points for parabolas.
• Rearrange the equation to '$x=$x= ....' and ask "Is there any $y$y value the equation cannot handle?"

 

 

When looking at parabolas, unless given a restricted domain, the domain is all real $x$x, also written as the set $\left(-\infty,\infty\right)$(−∞,∞). The range is either $\left[\text{min value},\infty\right)$[min value,∞) for concave up parabolas, or $\left(-\infty,\text{max value}\right]$(−∞,max value] for concave down parabolas. This is illustrated in the following image.  

Practice questions

Question 3

Question 4

question 5

Question 6

 

Co-domain and mapping

The co-domain for a function is related to the range but can be thought of as all the possible outputs for a given function. In contrast, the range is the actual set of outputs for the given function.

For example, consider the function $f(x)=3x$f(x)=3x for all positive integers $x$x. It would be reasonable to define the co-domain as all positive integers as well. However, the outputs will actually only be multiples of $3$3. So the range is all multiples of $3$3, for example, $3,6,9,\ldots$3,6,9,…

The co-domain is included in the notation of the function when using arrow notation.

If a function $f$f has the domain $X$X and co-domain $Y$Y, the following notation may be used:

$f$f  :  $X\rightarrow Y$X→Y

This reads as "the function $f$f from $X$X to $Y$Y" or "the function $f$f mapping the elements of the set $X$X to the elements of the set $Y$Y".

 

Practice questions

QUESTION 7