2.6 – Functions

Functions

A mathematical relation is a set of ordered pairs $(x,y)$ . The relation describes which $x$ and $y$ values can be paired together. The set of all $x$ values in the relation is called the domain, and the set of all $y$ values in the relation is called the range.

Identify the domain and range of the set of ordered pairs
$\{(5,6),(3,2),(5,4),(1,9),(8,12),(3,9)\}$ The domain is the set of $x$ values, and the range is the set of $y$ values, so we have: Domain = $\{1,3,5,8\}$ Range = $\{2,4,6,9,12\}$

Identify the domain and range of the set of ordered pairs

$\{(5,6),(3,2),(5,4),(1,9),(8,12),(3,9)\}$

The domain is the set of $x$ values, and the range is the set of $y$ values, so we have:

Domain = $\{1,3,5,8\}$

Range = $\{2,4,6,9,12\}$

Identify the domain and range of the relation
$\{(1,21),(4,17),(7,8),(9,25),(13,5),(16,26),(18,8),(24,12)\}$ Domain = $\{1,4,7,9,13,16,18,24\}$ Range = $\{5,8,12,17,21,25,26\}$

Identify the domain and range of the relation

$\{(1,21),(4,17),(7,8),(9,25),(13,5),(16,26),(18,8),(24,12)\}$

Domain = $\{1,4,7,9,13,16,18,24\}$

Range = $\{5,8,12,17,21,25,26\}$

A function is a special relation. It is a relation for which any particular $x$ value is paired only with one particular $y$ vallue. That is, a function is a set of ordered pairs $(x,y)$ for which every $x$ value is paired with exactly one $y$ value. That is, a function cannot have two ordered pairs with the same $x$ -coordinate, but different $y$ -coordinates.

Determine if the relation is a function
$\{(1,2),(3,2),(4,7),(8,9)\}$ This relation is indeed a function, because every $x$ value is paired up with only one $y$ value.

Determine if the relation is a function

$\{(1,2),(3,2),(4,7),(8,9)\}$

This relation is indeed a function, because every $x$ value is paired up with only one $y$ value.

Determine if the relation is a function
$\{(1,1),(2,6),(5,8),(1,7),(8,2)\}$ This relation is not a function, as the $x$ value of 1 is paired with two different $y$ values, 1 and 7

Determine if the relation is a function

$\{(1,1),(2,6),(5,8),(1,7),(8,2)\}$

This relation is not a function, as the $x$ value of 1 is paired with two different $y$ values, 1 and 7

When an equation is a function, it is often expressed in function notation, that is, the output variable $y$ is replaced with $f(x)$ so that is easily recognized as being a function. $f(x)$ is read “ $f$ of $x$ ” and indicates that the function has been named “ $f$ ” and its input variable is $x$ . A function need not necessarily always be named $f$ . When working with multiple functions, it is common to name them $f(x)$ , $g(x)$ , $h(x)$ and so forth. This helps to keep from using one function when another was intended. An example of a function in function notation:

$f(x)=2x+5$

The domain of a function equation is the set of all numbers that can be input into the function, and receive an output. That is, it is any number for which the function is defined. To find the domain of a function equation (generally referred to as simply a function from here on out), it is often easier to find the values of $x$ that cannot be input into the function.

Find the domain of the function $f(x)=2x$
We can evaluate numbers one by one, but that is certainly not a good method to check an infinite number of possible input values. Instead, we reason that any number can be doubled, so any number can be input into the equation. The domain of this function then, is all real numbers. In symbols, we write this as $x\in\mathbb{R}$ This means that $x$ is an element of $\mathbb{R}$ , which is the symbol for all “Real numbers”. It is the same as the interval notation $(-\infty,\infty)$

Find the domain of the function $f(x)=2x$

We can evaluate numbers one by one, but that is certainly not a good method to check an infinite number of possible input values. Instead, we reason that any number can be doubled, so any number can be input into the equation. The domain of this function then, is all real numbers. In symbols, we write this as
$x\in\mathbb{R}$
This means that $x$ is an element of $\mathbb{R}$ , which is the symbol for all “Real numbers”. It is the same as the interval notation $(-\infty,\infty)$

Some functions have restrictions that cane be found quickly by understanding their behavior.

Find the domain of the function $g(x)=\sqrt{x+5}$
First, by knowing the properties of the square root, our job is made much easier. We know that we can take the square root of a positive number, and we can take the square root of zero (the square root of 0 is 0). But we cannot take the square root of a negative number. So the radicand (the number, or expression underneath the square root function) must be at least zero. As we learned earlier, this means that we have the inequality: $x+5\ge0$ If we solve this inequality, we find our domain! $x+5\ge5$ $x\ge-5$ To write this in interval notation: $x\in[-5,\infty)$

Find the domain of the function $g(x)=\sqrt{x+5}$

First, by knowing the properties of the square root, our job is made much easier. We know that we can take the square root of a positive number, and we can take the square root of zero (the square root of 0 is 0). But we cannot take the square root of a negative number. So the radicand (the number, or expression underneath the square root function) must be at least zero. As we learned earlier, this means that we have the inequality:

$x+5\ge0$

If we solve this inequality, we find our domain!

$x+5\ge5$
$x\ge-5$

To write this in interval notation:

$x\in[-5,\infty)$

Find the domain of the function $h(x)=\sqrt{15-3x}$
Again, the radicand must be greater than or equal to zero: $15-3x\ge0$ $-3x\ge-15$ $x\le5$ So $x\in(-\infty,5]$

Find the domain of the function $h(x)=\sqrt{15-3x}$

Again, the radicand must be greater than or equal to zero:

$15-3x\ge0$
$-3x\ge-15$
$x\le5$

So $x\in(-\infty,5]$

Another function to consider is a fraction with the variable $x$ in the denominator. To deal with these functions, we recall that one can not divide by zero, so our domain is restricted and can not include any $x$ value that creates a zero in the denominator

Find the domain of the function $f(x)=\dfrac{3}{x}$
We can set up our restriction as in inequality using the “not equal to” sign, $\not=$ . Here, the denominator is simply the variable $x$ , so we have $x\not=0$ And there is nothing more to solve. $x$ simply can not equal zero. But it can be anything else! This means that we have a union. All of the numbers less than 0 and all of the numbers greater than zero. To write this in interval notation: $x\in(-\infty,0)\cup(0,\infty)$

Find the domain of the function $f(x)=\dfrac{3}{x}$

We can set up our restriction as in inequality using the “not equal to” sign, $\not=$ . Here, the denominator is simply the variable $x$ , so we have
$x\not=0$

And there is nothing more to solve. $x$ simply can not equal zero. But it can be anything else! This means that we have a union. All of the numbers less than 0 and all of the numbers greater than zero. To write this in interval notation:

$x\in(-\infty,0)\cup(0,\infty)$

Find the domain of the function $g(x)=\dfrac{27}{2x-8}$
We again will set the denominator not equal to zero: $2x-8\not= 0$ And solve for $x$ : $2x\not=8$ $x\not=4$ $x\in(-\infty,4)\cup(4,\infty)$

Find the domain of the function $g(x)=\dfrac{27}{2x-8}$

We again will set the denominator not equal to zero:
$2x-8\not= 0$
And solve for $x$ :
$2x\not=8$
$x\not=4$

$x\in(-\infty,4)\cup(4,\infty)$

Find the domain of the function $h(x)=\dfrac{7}{3x-2}$
Again, we set the denominator not equal to zero, and then solve: $3x-2\not=0$ $3x\not=2$ $x\not=\frac{2}{3}$ $x\in\left(-\infty,\frac{2}{3}\right)\cup\left(\frac{2}{3},\infty\right)$

Find the domain of the function $h(x)=\dfrac{7}{3x-2}$

Again, we set the denominator not equal to zero, and then solve:
$3x-2\not=0$
$3x\not=2$
$x\not=\frac{2}{3}$

$x\in\left(-\infty,\frac{2}{3}\right)\cup\left(\frac{2}{3},\infty\right)$

Find the domain of the function $f(x)=\dfrac{2}{\sqrt{6-2x}}$
In this problem, we have two issues. First, we have a denominator with a variable in it, and second, the denominator has a square root! We can reason our way through the domain restrictions here, noting that the radicand must be greater than or equal to zero. However, the denominator itself can not be zero, so we simply need to remove the equality, and we have our restriction: $6-2x>0$ $-2x>-6$ $x<3$ $x\in(-\infty,3)$

Find the domain of the function $f(x)=\dfrac{2}{\sqrt{6-2x}}$

In this problem, we have two issues. First, we have a denominator with a variable in it, and second, the denominator has a square root! We can reason our way through the domain restrictions here, noting that the radicand must be greater than or equal to zero. However, the denominator itself can not be zero, so we simply need to remove the equality, and we have our restriction:
$6-2x>0$
$-2x>-6$
$x<3$

$x\in(-\infty,3)$

To evaluate a function for a particular value, the $x$ is replaced with the value and then the expression is simplified. The final result is the corresponding $y$ value to the $x$ value input. To express that we are evaluating a function $f(x)$ for a value, for example 3, we write $f(3)$ . This is read “ $f$ of 3” or equivalently, “the function evaluated at 3”.

If $f(x)=3x+1$ , find $f(2)$ and $f(5)$
$\begin{align} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f(2)&=3(2)+1& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f(5)&=3(5)+1\\ &=6+1& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ &=15+1\\ &=7& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ &=16 \end{align}$ Thus, we have the ordered pairs (2,7) and (5,16)

If $f(x)=3x+1$ , find $f(2)$ and $f(5)$

$\begin{align*} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f(2)&=3(2)+1& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f(5)&=3(5)+1\\ &=6+1& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ &=15+1\\ &=7& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ &=16 \end{align*}$

Thus, we have the ordered pairs (2,7) and (5,16)

If $g(x)=3x^2+2x-4$ , find $g(0)$ , $g(-2)$ and $g(1)$
$\begin{align}g(0)=&3(0)^2+2(0)-4\\&=3(0)+2(0)-4\\&=0+0-4\\&=-4\end{align}$ $\begin{align}g(-2)=&3(-2)^2+2(-2)-4\\&=3(4)+2(-2)-4\\&=12-4-4\\&=4\end{align}$ and $\begin{align}g(1)=&3(1)^2+2(1)-4\\&=3(1)+2(1)-4\\&=3+2-4\\&=1\end{align}$ This gives us the ordered pairs (0,-4), (-2,8) and (1,1)

If $g(x)=3x^2+2x-4$ , find $g(0)$ , $g(-2)$ and $g(1)$

$\begin{align*}g(0)=&3(0)^2+2(0)-4\\&=3(0)+2(0)-4\\&=0+0-4\\&=-4\end{align*}$

$\begin{align*}g(-2)=&3(-2)^2+2(-2)-4\\&=3(4)+2(-2)-4\\&=12-4-4\\&=4\end{align*}$

and

$\begin{align*}g(1)=&3(1)^2+2(1)-4\\&=3(1)+2(1)-4\\&=3+2-4\\&=1\end{align*}$

This gives us the ordered pairs (0,-4), (-2,8) and (1,1)

PreCalculus I

2.6 – Functions

Functions

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