1. Functions

Lesson

A relation is a set of ordered pairs which represent a relationship. For example, think of the names of the people in your class and their heights. If I gave you a height (e.g. $162$162 cm), you could tell me all the names of the people who are this tall and there may be more than one person. Let's say someone came to your class looking for the person who was $162$162 cm tall, that description might fit four people! There's not one clear answer. This data could be expressed as a relation.

A function is a relationship between two quantities in which every input corresponds to exactly **one **output. Recall the dependent and independent variables that are present in equations. Functions work in a similar way, where the way the dependent variable, $f(x)$`f`(`x`), varies depending on the rule that is applied to the independent variable, $x$`x`. For example, if we think of a vending machine selling juice, the amount of money we have to pay (the dependent variable) depends on the amount of juice we want to buy (the independent variable). Let's say each bottle of juice cost $\$3$$3. If we bought one bottle, it would cost $\$3$$3, if we bought two bottles, it would cost $\$6$$6 and so on. Do you notice how the value of our independent variable (the number of bottles of juice) always produces a different, unique dependent variable (cost)? This is an example of a function.

Let's look at another example. Say we have the expression $f(x)=2x$`f`(`x`)=2`x`. Let's construct a table of values to record the results:

$x$x |
$-1$−1 | $0$0 | $1$1 | $2$2 |
---|---|---|---|---|

$f(x)$f(x) |
$-2$−2 | $0$0 | $2$2 | $4$4 |

See how each $x$`x` value gives a unique $f(x)$`f`(`x`) value? This means this data displays a function.

Remember!

For the function $f(x)$`f`(`x`),

If $x$`x` is the input, then $f(x)$`f`(`x`) is the output and gives us the ordered pair $\left(x,f(x)\right)$(`x`,`f`(`x`)).

For example, when I substitute in $x=2$`x`=2 and get $y=5$`y`=5, we can write $f(2)=5$`f`(2)=5, and this can be represented with the ordered paid $\left(2,5\right)$(2,5).

Recall that the set of all possible input (independent) values of a relation is called the domain and the set of all possible output (dependent) values of a relation is called the range.

From a table of values, we can determine if a relation is also a function by looking at the values of the independent variable. If there are any points that have the **same **$x$`x`-value, but **different **$y$`y`-values, then that relation is **not **a function.

If you can draw a vertical line anywhere on a graph so that it crosses the graph in more than one place, then the relation is **not **a function.

Here is an example of a relation that is not a function. See how when I drew in the blue vertical line, it crossed the graph in two places?

In other words, functions have to pass the vertical line test **at every point. **

Here is one example of a function.

Here is another function.

If you can write a relationship between $x$`x` and $y$`y` then we can see that there is a relation. However, if this relationship only yields one value of $y$`y` for each $x$`x` value, then it is a function. We will start by writing the relationship in the form $y=...$`y`=....

When we solve for $y$`y` in $2y-4x=10$2`y`−4`x`=10, we can tell whether it's a function or a relation.

$2y-4x=10$2`y`−4`x`=10

$2y=4x+10$2`y`=4`x`+10

$y=2x+5$`y`=2`x`+5

See that each value of $x$`x` only yields one $y$`y` value, which means that it is a **function, **so we can write $f(x)=2x+5$`f`(`x`)=2`x`+5.

Let's try this process for the equation $y^2=x$`y`2=`x`:

$y=\pm\sqrt{x}$`y`=±√`x`

See how $y$`y` could be $\sqrt{x}$√`x` or $-\sqrt{x}$−√`x`? Since there are two possible values of $y$`y`, we can only say it is a **relation**.

Remember!

While all functions are relations, not all relations are functions.

For the set of points $\left\{\left(1,5\right),\left(1,1\right),\left(7,-2\right),\left(-5,-10\right)\right\}${(1,5),(1,1),(7,−2),(−5,−10)},

**a)** List the domain.

**Think:** The domain is the set of all possible values of the independent variable. When writing a set, we only have to write each value once, so we won't list $1$1 twice even though it appears in $\left(1,5\right)$(1,5) and $\left(1,1\right)$(1,1).

**Do:** The domain is the set $\left\{1,7,-5\right\}${1,7,−5}

**b)** List the range.

**Think:** The range is the set of all possible values of the dependent variable.

**Do:** The range is the set $\left\{5,1,-2,-10\right\}${5,1,−2,−10}

**c) **Determine whether this set represents a relation, a function or both.

**Think:** Does each $x$`x` value have a unique $y$`y` value?

**Do:** There are $2$2 possible $y$`y` values when $x=1$`x`=1. Therefore this describes only a relation, **not **a function.

Determine whether this set represents a relation, a function or both. $\left\{\left(1,5\right),\left(7,-2\right),\left(-5,-10\right),\left(13,-13\right)\right\}${(1,5),(7,−2),(−5,−10),(13,−13)}.

**Think:** Does each $x$`x` value have a unique $y$`y` value?

**Do:** In this set of coordinates, each $x$`x` value has only one unique $y$`y` value. Therefore this describes a function (which is also a relation).

Determine whether the following graph describes a function or a relation.

**Do:** This graph passes the vertical line test. Therefore, it is a function (and is also a relation).

Determine whether the following graphs describe relations, and whether they describe functions.

- Loading Graph...
Select all answers that apply.

Function

ARelation

BFunction

ARelation

B - Loading Graph...
Select all answers that apply.

Function

ARelation

BFunction

ARelation

B

The pairs of values in the table represent a relation between $x$`x` and $y$`y`. Do they represent a function?

$x$x |
$-8$−8 | $-7$−7 | $-6$−6 | $-3$−3 | $2$2 | $7$7 | $9$9 | $9$9 | $10$10 |
---|---|---|---|---|---|---|---|---|---|

$y$y |
$8$8 | $13$13 | $-18$−18 | $-16$−16 | $-15$−15 | $-2$−2 | $-4$−4 | $11$11 | $-9$−9 |

Yes

ANo

BYes

ANo

B

Consider the function $f\left(x\right)=4+x^3$`f`(`x`)=4+`x`3.

Evaluate $f\left(4\right)$

`f`(4).Evaluate $f\left(-2\right)$

`f`(−2).