Common Core Algebra 2 - 2020 Edition
1.04 Domain and range and continuity
Lesson

## Domain and range

We looked at functions and relations as a relationship between two quantities, usually written using $x$x's and $y$y's as ordered pairs. We also learned that relations can only be functions if there's 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. We also know that graphically, we can use the vertical line test to see if a relation is a function.

The concept of domain and range is very important and will keep coming up in our studies of mathematics.

Remember!

Domain - the set of all possible values of the independent variable ($x$x-values) in a relation

Range - the set of all possible values of the dependent variable ($y$y-values) in a relation

There are a number of ways to find the domain and range of a relation.

One is to look at the coordinates given and simply list the possible values as a set. For example, for the relation $\left\{\left(1,2\right),\left(5,3\right),\left(2,-7\right),\left(5,-1\right)\right\}${(1,2),(5,3),(2,7),(5,1)}, 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 we can write the domain as $-1\le x\le1$1x1. Similarly, the graph goes vertically from $-2$2 to $2$2 so the range can be written as $-2\le y\le2$2y2.

#### Worked example

##### question 1

A cafe is running a promotion where every $3$3 cups of coffee earns you a free cup, and each cup costs $\$3.503.50

a) Make a table of prices for $1$1 to $8$8 cups of coffee

Think: Which pairs of numbers of cups would be equal to the same price?

Do:

$3$3 cups and $4$4 cups would cost the same as the $4$4th cup would be free, same goes for $7$7 and $8$8 cups. Therefore:

Number of cups Price ($)$1$1$3.50$3.50$2$2$7$7$3$3$10.50$10.50$4$4$10.50$10.50$5$5$14$14$6$6$17.50$17.50$7$7$21$21$8$8$21$21 b) Is this relation a function? Think: In word problems,$x$x usually represents the independent variable and$y$y the dependent variable. Are there any$x$x values here with more than one$y$y value? Do: The left column can be represented by$x$x and the right by$y$y. We can see that we can some doubling up of$y$y values for different$x$x values but for each$x$x value there is only one$y$y value. Therefore this is a function. #### Practice questions ##### Question 2 The function$f\left(x\right)=\sqrt{x+1}$f(x)=x+1 has been graphed. Loading Graph... 1. State the domain of the function. Express as an inequality. 2. Is there a value in the domain that can produce a function value of$-2$2? Yes A No B Yes A No B ##### question 3 The function$f$f is used to determine the area of a square given its side length. 1. Which of the following values is not part of the domain of the function?$-8$8 A$6.5$6.5 B$9$9$\frac{1}{3}$13 C$\sqrt{78}$78 D$-8$8 A$6.5$6.5 B$9$9$\frac{1}{3}$13 C$\sqrt{78}$78 D 2. For$n\ge0$n0, state the area function for a side length of$n$n. 3. Plot the graph of the function$f$f. Loading Graph... ## Interval Notation An interval on the real number line is the set of numbers between two endpoints. One, both, or neither endpoints can belong to the interval and there are notations for each possibility. We have already seen inequality notation for intervals, for example$-33<x<6 which does not include either endpoint, $-3\le x<6$3x<6 which includes on endpoint, and $-3\le x\le6$3x6 which includes both endpoints. We now want to look at a more succinct notation called interval notation.

Let's start with intervals that do not include their endpoints such as $-33<x<6. We call this an open interval. The notation for these uses rounded parentheses. For example, we write$\left(-3,6\right)$(3,6) to mean the set of numbers between$-3$3 and$6$6 but not including either$-3$3 or$6$6 Intervals that include their endpoints are called closed intervals, for example$-3\le x\le6$3x6. To specify a closed interval including all numbers between$-3$3 and$6$6, for example, we write$\left[-3,6\right]$[3,6]. The square parentheses indicate that the numbers$-3$3 and$6$6 are considered to belong to the interval. It is also possible for an interval to be closed at one end but open at the other. For example,$[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]$(,b] Thus, for example, we can indicate the whole real number line with the notation$\left(-\infty,\infty\right)$(,). We use this notation for domain and range. To state a domain, we may want to say that$x$x is in a certain interval, we can write$x\in(a,b)$x(a,b). We understand this to mean that$x$x is greater than$a$a and$x$x is less than$b$b. That is, this is the same as writing$aa<x<b, meaning that $x$x is strictly between $a$a and $b$b.

Notation Inequality Interval Graphical
Not including $<$< or $>$> $($( or $)$) open dot (hollow)
Including $\le$ or $\ge$ $[$[ or $]$] closed dot (filled in)

To join together two intervals, we can use the union symbol, $\cup$. For example, $x\in(-\infty,5)\cup(5,\infty)$x(,5)(5,) is all real numbers except $5$5

#### Worked examples

##### Question 4

Convert the following from inequality notation to interval notation.

a) $-22<x10 Think: If an endpoint is included, we use a square bracket. If an endpoint is not included we use a round bracket. Do:$-22<x10 can also be written as $(-2,10]$(2,10]

b) $y\ge3$y3

Think: This means all $y$y-values which are greater than or equal to $3$3, so from $3$3 to infinity.

Do: $y>3$y>3 can also be written as $[3,\infty)$[3,)

##### Question 5

State the domain of the interval below using interval notation.

Think: Each tick is representing $0.5$0.5, so the closed dot is at $-3$3 and the open dot is at $1$1. We want to include$-3$3, but not $1$1. We always start with the smaller number.

Do: This interval can be represented as $[-3,1)$[3,1).

##### Question 6

State the domain of the interval below using interval notation.

Think: Each tick is representing $0.5$0.5, so the closed dot is at $-1$1 and it continues off to infinity.

Do: This interval can be represented as $[-1,\infty)$[1,).

##### Question 7

State the domain of the interval below using interval notation.

Think: Each tick is representing $0.5$0.5, so we have an open dot at $-1$1 and it continues down to $-\infty$ and an open dot at $1$1 and it continues up to $\infty$. We will need to join these two pieces using the union symbol.

Do: This domain can be represented as $(-\infty,-1)\cup(1,\infty)$(,1)(1,).

Careful!

When using interval notation, be sure to always have the smaller value first. For example, it must be $(-\infty,-1)$(,1), not $(-1,-\infty)$(1,)

#### Practice questions

##### Question 8

Consider the pictured inequality.

1. The endpoints of this interval are $\editable{}$ and $\editable{}$.

2. Which of the following is the correct notation for the pictured interval?

$[$[$-1$1,$2$2$]$]

A

$($($-1$1,$2$2$]$]

B

$[$[$-1$1,$2$2$)$)

C

$($($-1$1,$2$2$)$)

D

$[$[$-1$1,$2$2$]$]

A

$($($-1$1,$2$2$]$]

B

$[$[$-1$1,$2$2$)$)

C

$($($-1$1,$2$2$)$)

D

##### Question 9

Consider the curve on the graph below.

1. State the domain of the function using interval notation.

2. State the range of the function using interval notation.

## Continuity

Recall that the domain of a function is the set of $x$x values over which the function is defined. Graphically, we can think of the domain as all values of $x$x which correspond to a point on the curve.

A function $f\left(x\right)$f(x) is said to be continuous at a point $x=c$x=c if:

i. the function is defined at that point (that is, $c$c is in the domain of $f\left(x\right)$f(x)), and

ii. we can get function values as close as we like to $f\left(c\right)$f(c) by taking a small enough region of domain values around $x=c$x=c.

A function $f\left(x\right)$f(x) is then said to be a continuous function if it is continuous at all points in its domain. Similarly, a function is continuous over an interval if it is continuous at all points in that interval. You can think of it as being able to draw that interval of the function without lifting your pencil.

If instead there exists a point in the function's domain at which it is not continuous, we say that the function is discontinuous at that point.

Let's now take a look at some different types of continuous functions by inspecting their graphs.

### Continuous functions over the real numbers

Linear functions are defined for all real numbers. We can represent this by saying its domain is the interval $\left(-\infty,\infty\right)$(,). It is also continuous over this domain, so it is a continuous function.

A linear function

Look at the linear function in the graph above. At the domain value $x=2$x=2 we have the function value $f\left(2\right)=3$f(2)=3. We can get other function values as close as we like to $3$3 by taking a small enough region of the domain around $x=2$x=2. Watch the applet below for a demonstration.

 Created with Geogebra

There is nothing special about $x=2$x=2 in this case - the same is true for any other point in the domain! So the function above is a continuous function, as are all linear functions.

Other examples of continuous functions which have a domain of $\left(-\infty,\infty\right)$(,) include any type of polynomial function (quadratic functions, cubic functions etc) as well as exponential functions.

Some of these types are sketched below:

A cubic function

An exponential function

### Continuous functions over smaller domains

Many functions are not defined for every real number, but are still continuous at every point in their domain. Let's take a look at some of these types of functions.

A rational function has a form such as $f\left(x\right)=\frac{a}{x-h}+k$f(x)=axh+k. Since $f(x)$f(x) is not defined for $x=h$x=h, in particular it has a vertical asymptote at $x=h$x=h, it can't be continuous at $x=h$x=h. What about over its domain though?

A rational function

The rational function in the graph above has a vertical asymptote at $x=1$x=1. The domain of this function is all of the real numbers except for $x=1$x=1, which can be written as $\left(-\infty,1\right)\cup\left(1,\infty\right)$(,1)(1,). Now if we exclude $x=1$x=1 from consideration, we can see that the function is continuous over each value in the domain - even if we take an $x$x-value very close to $x=1$x=1, we can still move along the curve (even just a little bit!) in either direction. This means that the rational function is a continuous function over its domain.

A square root function, which takes a form such as $f\left(x\right)=a\sqrt{x-h}+k$f(x)=axh+k, is defined when the input of the square root is not negative, so the domain is $x\ge h$xh or $[h,\infty)$[h,).

A square root function

The function $f\left(x\right)=\sqrt{x+4}-1$f(x)=x+41 has been graphed above. The value under the square root cannot be negative, and (as we can see from the graph) this results in a domain of $\left[-4,\infty\right)$[4,). The function is continuous at all points in this domain, and so it is also a continuous function over its domain.

### Types of discontinuities

There are three main types of discontinuities that prevent functions from being continuous over the real number or possible even over their domain.

Asymptotes - vertical asymptotes typically occur when we are dividing by zero, so the function is not defined at that point, so is therefore not continuous at that point.

Example: $f(x)=\frac{1}{x-1}-2$f(x)=1x12 shown below. $f(1)$f(1) is not defined, so $f(x)$f(x) is discontinuous at $x=1$x=1.

Removable or point discontinuities - these discontinuities typically occur when we are dividing by zero, but the factor actually cancels out, so the function is not defined at that point, but there is not a vertical asymptote.

Example: $f(x)=\frac{x^2-5x-6}{x+1}$f(x)=x25x6x+1 shown below. $f(-1)$f(1) is not defined, so $f(x)$f(x) is discontinuous at $x=-1$x=1. This is not an asymptote because $f(x)$f(x) simplifies to $f(x)=x-6$f(x)=x6, $x\ne-1$x1 if we factor the numerator and cancel.

Jump - A jump discontinuity typically occur for piecewise functions that have two separate pieces, but the left side does not connect with the right side. when we are dividing by zero, so the function is not defined at that point, so is therefore not continuous at that point.

Example: See the piecewise function below. $f(0)$f(0) is not defined, so $f(x)$f(x) is discontinuous at $x=0$x=0.

### Discontinuous functions

As we have seen so far, most of the functions that we are familiar with are continuous over their domain. Not all functions are continuous though! Let's take a look at an example of a step function:

A step function

In the graph above, the step function has a step at $x=0$x=0. It takes a constant value (in this case $-1$1) for $x<0$x<0 and a different constant value (in this case $+1$+1) for $x\ge0$x0, and so its domain is $\left(-\infty,\infty\right)$(,) (that is, all of the real numbers).

This function has a point of discontinuity at $x=0$x=0 - no matter how small we make our region of domain values around $x=0$x=0, we will always include part of the function that takes the value $-1$1 and part of the function that takes the value $+1$+1. This means that the step function is not a continuous function.

Note that the function is still continuous over the interval $\left(0,\infty\right)$(0,) and over the interval $\left(-\infty,0\right)$(,0).

#### Practice questions

##### Question 10

Consider the function $f\left(x\right)=\frac{x^2-25}{x-5}$f(x)=x225x5 drawn below.

1. What is the domain of the function $f\left(x\right)=\frac{x^2-25}{x-5}$f(x)=x225x5?

$\left(-\infty,5\right)\cup\left(5,\infty\right)$(,5)(5,)

A

$\left(-\infty,\infty\right)$(,)

B

$\left[5,\infty\right)$[5,)

C

$\left(-\infty,5\right]$(,5]

D

$\left(-\infty,5\right)\cup\left(5,\infty\right)$(,5)(5,)

A

$\left(-\infty,\infty\right)$(,)

B

$\left[5,\infty\right)$[5,)

C

$\left(-\infty,5\right]$(,5]

D
2. Select the largest region over which the function is continuous.

$\left(-\infty,5\right)$(,5)

A

$\left(-\infty,5\right)\cup\left(5,\infty\right)$(,5)(5,)

B

$\left(-\infty,\infty\right)$(,)

C

$\left(5,\infty\right)$(5,)

D

$\left(-\infty,5\right)$(,5)

A

$\left(-\infty,5\right)\cup\left(5,\infty\right)$(,5)(5,)

B

$\left(-\infty,\infty\right)$(,)

C

$\left(5,\infty\right)$(5,)

D
3. So is the function $f\left(x\right)=\frac{x^2-25}{x-5}$f(x)=x225x5 a continuous function over its domain?

No

A

Yes

B

No

A

Yes

B

##### Question 11

Which of the following functions are not continuous?

Select all that apply.

A

B

C

D

A

B

C