We have been exploring how functions and relations link inputs and outputs through a given equation.
Domain and range give us a way to describe these sets of inputs and outputs.
The first way to find a relation's domain and range is to simply look at the coordinates given and list the possible values.
Consider the relation described by the set of points:$\left\{\left(1,2\right),\left(5,3\right),\left(2,-7\right),\left(5,-1\right)\right\}${(1,2),(5,3),(2,−7),(5,−1)}. For this relation, 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.
Another 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 we can write the domain as $-1\le x\le1$−1≤x≤1. Similarly, the graph goes vertically from $-2$−2 to $2$2 so the range can be written as $-2\le y\le2$−2≤y≤2.
We could write the domain for the example above as $x\ge-1$x≥−1 AND $x\le1$x≤1. which is technically correct but not as elegant as the compound inequality used above. The values of $x$x that satisfy this compound inequality include all numbers between $-1$−1 and $1$1 inclusive with the inequality showing that $x$x actually is between $-1$−1 and $1$1.
Our second option is to use interval notation for domain and range. An interval on the real number line is the set of numbers between two endpoints. One or both endpoints, or neither, can belong to interval and there are notations for each possibility.
Intervals that include their endpoints are called closed intervals. This is the case for the inequality $-1\le x\le1$−1≤x≤1. Closed intervals use square brackets, and so we could write this as $\left[-1,1\right]$[−1,1]. Intervals that do not include their endpoints are called open intervals. The notation for these uses rounded brackets. For example, we write $\left(0,100\right)$(0,100) to mean the set of numbers between $0$0 and $100$100 but not including either $0$0 or $100$100. It is also possible for an interval to be closed at one end but open at the other. For example, $[0,\sqrt{2})$[0,√2) or $(-9,0]$(−9,0].
An interval with no upper bound is indicated with the $\left[a,\infty\right)$[a,∞) sign. Such intervals are said to be open on the right. Similarly, an interval with no lower bound is open on the left and is notated with the sign $\left(-\infty,b\right]$(−∞,b].
Think about a straight line like $y=2x+1$y=2x+1, pictured below. Its domain and range have no restrictions at all. And so, we could write the domain and range in interval notation as $\left(-\infty,\infty\right)$(−∞,∞).
Use the following applet to explore the domain and range of linear and quadratic function using interval notation.
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A relation or function may be given to us as an equation rather than a graph. The equation may have implied restrictions because of the mathematical operations involved within it and this is called the natural domain. In general, the best way to approach these questions is to ask yourself "is there any input value the equation cannot handle?".
For example, the function $f\left(x\right)=\sqrt{x}$f(x)=√x cannot handle negative real numbers. Hence, its domain is: $x\ge0$x≥0. The function $f\left(x\right)=\frac{1}{x}$f(x)=1x cannot handle zero as we know it is impossible to divide any number by zero. Other than that however there aren't any restrictions. Hence, its domain is the compound interval: $\left(-\infty,0\right)\cup\left(0,\infty\right)$(−∞,0)∪(0,∞).
Note that we use the symbol '$\cup$∪' when we want to specify more than one interval that applies to a relation. This happens when functions or relations have discontinuities, usually because of asymptotes. You will see more of this symbol when you use set notation in the Probability topic!
Finally, it is important to consider how combining functions through addition, subtraction, multiplication or division will affect domain and range. The restrictions on each component will affect the combined irrespective of the new functions structure. For example, if $f(x)$f(x) has the domain $x>3$x>3 and $g(x)$g(x) has a domain $x\le10$x≤10, then $(f-g)(x)$(f−g)(x) will only exist for $3
Consider the relation in the table.
$x$x | $y$y |
---|---|
$7$7 | $1$1 |
$7$7 | $9$9 |
$8$8 | $3$3 |
$5$5 | $2$2 |
$3$3 | $6$6 |
What is the domain of the relation? Enter the values separated by commas.
What is the range of the relation? Enter your answers separated by commas.
Is this relation a function?
Yes
No
Consider the function that has been graphed.
What is the domain of the function?
$x\ge5$x≥5
$x>0$x>0
$x<-5$x<−5
all real $x$x
What is the range of the function? Give your answer as an inequality.
Consider the graph of the circle shown below.
State the domain of the graph in interval notation.
State the range of the graph in interval notation.
Consider the graph of the rational function $f\left(x\right)$f(x).
By filling in the gaps, determine the domain of $f\left(x\right)$f(x).
Domain of $f$f: $\left(-\infty,\editable{}\right)$(−∞,) $\cup$∪ $\left(\editable{},\editable{}\right)$(,) $\cup$∪ $\left(\editable{},\infty\right)$(,∞)
Which of the following is the range of $f\left(x\right)$f(x)?
$\left(-\infty,-6\right)\cup\left(-6,\infty\right)$(−∞,−6)∪(−6,∞)
$\left(-\infty,0\right)\cup\left(0,\infty\right)$(−∞,0)∪(0,∞)
$\left(-\infty,\infty\right)$(−∞,∞)
What is the domain of the function defined by $f\left(x\right)=\frac{1}{x+5}$f(x)=1x+5?
$\left(-\infty,1\right)$(−∞,1)$\cup$∪$\left(1,\infty\right)$(1,∞)
$\left(-\infty,0\right)$(−∞,0)$\cup$∪$\left(0,\infty\right)$(0,∞)
$\left(-\infty,-5\right)$(−∞,−5)$\cup$∪$\left(-5,\infty\right)$(−5,∞)
$\left(-\infty,5\right)$(−∞,5)$\cup$∪$\left(5,\infty\right)$(5,∞)
Consider the functions $f\left(x\right)=\frac{4x-3}{x+2}$f(x)=4x−3x+2 and $g\left(x\right)=\frac{x^2+8x+6}{x^2+7x+10}$g(x)=x2+8x+6x2+7x+10.
Find $\left(f+g\right)\left(x\right)$(f+g)(x), leaving the numerator in expanded form.
What is the domain of $\left(f+g\right)\left(x\right)$(f+g)(x)?
$\left(-\infty,-5\right)\cup\left(-5,-2\right)\cup\left(-2,\infty\right)$(−∞,−5)∪(−5,−2)∪(−2,∞)
$\left(-\infty,-5\right]\cup\left[-2,\infty\right)$(−∞,−5]∪[−2,∞)
$\left(-\infty,\infty\right)$(−∞,∞)
$\left(-\infty,-2\right)\cup\left(-2,\infty\right)$(−∞,−2)∪(−2,∞)