## Section5.3Introduction to Function Notation

##### Functions from a Graphical Perspective.

A function is a set of ordered pairs with the property that no two points in the set share the same first coordinate. Specifically, if the function is a set of points of form $(x,y)\text{,}$ we say that $y$ is a function of $x\text{,}$ and if the name of the function is $f\text{,}$ we write $y=f(x)\text{.}$ The symbols $y=f(x)$ are read aloud as "$y$ equals $f$ of $x$" or "$y$ equals $f$ at $x\text{.}$" The $y$-coordinate of he point is called the function value at the $x$-coordinate of the point.

Consider the function, $f\text{,}$ that consists of the ordered pairs shown in Table 5.3.1. From the given ordered we know each of the following function values.

\begin{align*} f(0)\amp=7\\ f(1)\amp=-3\\ f(2)\amp=11\\ f(3)\amp=7 \end{align*}

Using words, we say that the value of $f$ at $0$ is $7\text{,}$ the value of $f$ at $1$ is $-3\text{,}$ the value of $f$ at $2$ is $11\text{,}$ and the value of $f$ at $3$ is $7\text{.}$

Consider the function, $g\text{,}$ shown in Figure 5.3.2. From the points (ordered pairs) that have been stated, we can infer the following function values.

\begin{align*} g(-4)\amp=2\\ g(1)\amp=3\\ g(5)\amp=-1 \end{align*}

If we were asked to evaluate the function at $4\text{,}$ we would first locate the point with an $x$-coordinate of $4\text{.}$ Then the function value would be the corresponding $y$-coordinate. In this case the relevant point is $(4,0)$ and we would report the function value by writing $f(4)=0\text{.}$ You can use Figure 5.3.3 to explore this idea further.

Now suppose that we were asked to solve the equation $g(x)=4\text{.}$ Just like any other equation with the variable $x\text{,}$ we are looking for all values that we could substitute for $x$ that would result in a true statement. In words, the equation $g(x)=4$ could be stated as "determine the values of $x$ for which the value of the function$g$ is $2\text{.}$" Since the function value corresponds the to the $y$-coordinate, we can further refine the charge as follows: "locate the points on the curve $y=g(x)$ with $y$-coordinates equal to $4\text{,}$ then the solutions to $g(x)=4$ are the $x$-coordinates of those points"

In Figure 5.3.4 I've added the line $y=4$ to a graph of $y=g(x)\text{.}$ We can see that the line intersects the function $g$ at the points $(-3,4)$ and $(0,4)\text{,}$ so the solutions to the equation $g(x)=4$ are $-3$ and $4\text{.}$ Similarly, if we drew the line $y=1\text{,}$ we see that is would intersect $g$ at the points $(-4.5,1)$ and $(3,1)\text{,}$ so the solutions to the equation $g(x)=1$ are $-4.5$ and $1\text{.}$

Let's consider the equation $g(x)=-8\text{.}$ We can't draw the line $y=-8$ onto Table 5.3.5, nor can we see a point with an $y$-coordinate of $10\text{,}$so we'll have to take a different tact. The arrow on the left side of $g$ implies that $g$ continues forever in the leftward direction with the same slope. So one strategy we can employ is to use that slope to make a table for $g$ until we find the point that has a $y$-coordinate of $-8\text{.}$ I've done that in Table 5.3.5, and we can see that the relevant ordered pair is $(-8,-9)\text{.}$ From this we know that the solution to the equation $g(x)=-8$ is $-9\text{.}$

##### Function Formulas.

Functions are frequently described by a function formula. For example, consider

\begin{equation*} f(x)=x^2-8\text{.} \end{equation*}

Suppose that we want to determine the value of $f$ when $x$ has a value of $-3\text{.}$ To make this determination we need to replace $x$ with $-3$ on both sides of the equation and then simplify the expression on the right-side of the equation (AKA "do the math"). Let's do it.

\begin{align*} f(\highlight{x})\amp=\highlight{x}^2-\highlight{x}-8\\ f(\highlight{-3})\amp=(\highlight{-3})^2-(\highlight{-3})-8\\ \amp=9+3-8\\ \amp=4 \end{align*}

So the value of $f$ at $-3$ is $4\text{.}$

You can use Figure 5.3.6 to further explore the replacement of $x$ with a numerical value. Notice that the value of $x$ changes on both sides of the equal sign.

Let's see several more examples.

###### Example5.3.7.

Determine the value of $g(x)=\sqrt{9-2x}$ at $-8\text{.}$

Solution
\begin{align*} g(\highlight{x})\amp=\sqrt{9-2\highlight{x}}\\ g(\highlight{-8})\amp=\sqrt{9-2(\highlight{-8})}\\ \amp=\sqrt{9+16}\\ \amp=\sqrt{25}\\ \amp=5 \end{align*}

The value of $g$ at $-8$ is $5\text{.}$

###### Example5.3.8.

Let $f(x)=8-5x\text{.}$ Determine the value(s) of $x$ where $f$ has a value of $53\text{.}$

Solution

In this example we are being told that

\begin{equation*} f(x)=53 \end{equation*}

and being asked to determine the value of $x$ when this happens. That is, we are being asked to solve the equation

\begin{equation*} 8-5x=53\text{.} \end{equation*}

Let's get to solving.

\begin{align*} 8-5x\amp=53\\ 8-5x\subtractright{8}\amp=53\subtractright{8}\\ -5x\amp=45\\ \divideunder{-5x}{-5}\amp=\divideunder{45}{-5}\\ x\amp=-9 \end{align*}

The function, $f\text{,}$ has a value of $53$ when the value of $x$ is $-9$

###### Example5.3.9.

Determine the values of $g(5)$ and $g(3)$ where $g(x)=\frac{x+12}{x^2-4x+3}\text{.}$

Solution
\begin{align*} g(\highlight{x})\amp=\frac{\highlight{x}+12}{\highlight{x}^2-4\highlight{x}+3}\\ g(\highlight{5})\amp=\frac{\highlight{5}+12}{\highlight{5}^2-4(\highlight{5})+3}\\ \amp=\frac{17}{25-20+3}\\ \amp=\frac{17}{8} \end{align*}
\begin{align*} g(\highlight{x})\amp=\frac{\highlight{x}+12}{\highlight{x}^2-4\highlight{x}+3}\\ g(\highlight{3})\amp=\frac{\highlight{3}+12}{\highlight{3}^2-4(\highlight{3})+3}\\ \amp=\frac{15}{9-12+3}\\ \amp=\frac{15}{0} \end{align*}

In the parlance of Scooby Doo, Rut Roh! Division by zero never leads to good results. Put another way, no number is equal to $\frac{15}{0}\text{.}$ When an error like this arises, we say that the function value is "undefined." Or we might say that the function value "does not exist." Any way we put it, $g(3)$ is a non-thing (in this problem).

###### Example5.3.10.

Let $h(t)=\frac{2t+1}{t-6}\text{.}$ Determine each of the following.

1. The value(s) of $t$ where the value of $h$ is equal to $\frac{7}{2}\text{.}$

2. The value of $h$ when the value of $t$ is $5\text{.}$

3. Any and all values of $t$ where the function is undefined.

Solution
1. We are being asked to determine where $h(t)=\frac{7}{2}\text{.}$ To do so, we need to set the formula for $h(t)$ equal to $\frac{7}{2}$ and solve the resultant equation for $t\text{.}$

\begin{align*} \frac{2t+1}{t-6}\amp=\frac{7}{2}\\ 2(2t+1)\amp=7(t-6)\,\,\,\,\text{(I cross-multiplied.)}\\ 4t+2\amp=7t-42\\ 4t+2\subtractright{2}-\subtractright{7t}\amp=7t-42\subtractright{2}-\subtractright{7t}\\ -3t\amp=40\\ \divideunder{-3t}{-3}\amp=\divideunder{-44}{-3}\\ t\amp=\frac{44}{3} \end{align*}

The value of $h$ is $\frac{7}{2}$ when the value of $t$ is $\frac{44}{3}\text{.}$

2. We are being asked to determine $h(5)\text{.}$

\begin{align*} h(\highlight{t})\amp=\frac{2\highlight{t}+1}{\highlight{t}-6}\\ h(\highlight{5})\amp=\frac{2(\highlight{5})+1}{\highlight{5}-6}\\ \amp=\frac{11}{-1}\\ \amp=-11 \end{align*}

The value of $h$ at $5$ is $-11\text{.}$

3. We are essentially asked to determine what values of $t$ will make the formula

\begin{equation*} \frac{2t+1}{t-6} \end{equation*}

go haywire. This will only occur if we try to force division by zero, which in turn will only occur if we replace $t$ with $6\text{.}$ So the only value where $h$ is undefined is $6\text{.}$

This idea of determining where a function is or is not defined is explored in more detail in the section titled "Domain and Range".

### ExercisesExercises

Determine each of the stated function values.

###### 1.

Determine $g(7)$ where $g(t)=8-4t\text{.}$

Solution

\begin{aligned}[t] g(\highlight{7})\amp=8-4(\highlight{7})\\ \amp=8-28\\ \amp=-20 \end{aligned}

###### 2.

Determine $f(-3)$ where $f(x)=-x^2+5x+12\text{.}$

Solution

\begin{aligned}[t] f(\highlight{-3})\amp=-(\highlight{-3})^2+5(\highlight{-3})+12\\ \amp=-9-15+12\\ \amp=-12 \end{aligned}

###### 3.

Determine $h(33)$ where $h(x)=-\sqrt{\frac{x-25}{2}}\text{.}$

Solution

\begin{aligned}[t] h(\highlight{33})\amp=-\sqrt{\frac{\highlight{33}-25}{2}}\\ \amp=-\sqrt{\frac{8}{2}}\\ \amp=-\sqrt{4}\\ \amp=-2 \end{aligned}

###### 4.

Determine $t(22)$ where $t(x)=17\text{.}$

Solution

\begin{aligned}[t] t(\highlight{22})\amp=17\\ \end{aligned}

###### 5.

Determine $y(-6)$ where $y(t)=\abs{-9-t}+3\text{.}$

Solution

\begin{aligned}[t] y(\highlight{-6})\amp=\abs{-9-(\highlight{-6})}+3\\ \amp=\abs{-3}+3\\ \amp=3+3\\ \amp=6 \end{aligned}

###### 6.

Determine $z(-1)$ where $z(x)=\frac{x+6}{x^2+1}\text{.}$

Solution

\begin{aligned}[t] z(\highlight{-1})\amp=\frac{\highlight{-1}+6}{(\highlight{-1})^2+1}\\ \amp=\frac{5}{2} \end{aligned}

###### 7.

Determine $k(4)$ where $k(t)=3\sqrt{21-t^2}\text{.}$

Solution

\begin{aligned}[t] k(\highlight{4})\amp=3\sqrt{21-\highlight{4}^2}\\ \amp=3\sqrt{21-16}\\ \amp=3\sqrt{5} \end{aligned}

Determine the solution set for each of the following equations.

###### 8.

Solve $f(x)=12$ where $f(x)=4-x\text{.}$

Solution
\begin{align*} f(x)\amp=12\\ 4-x\amp=12\\ 4-x\subtractright{4}\amp=12\subtractright{4}\\ -x\amp=8\\ \multiplyleft{-1}-x\amp=\multiplyleft{-1}8\\ x\amp=-8 \end{align*}

The solution set is $\{-8\}\text{.}$

###### 9.

Solve $h(t)=100$ where $h(t)=t^2-21\text{.}$

Solution

The solution set is $\{-11,11\}\text{.}$

###### 10.

Solve $p(x)=-14$ where $p(x)=6-\abs{x}\text{.}$

Solution
\begin{align*} p(x)\amp=-14\\ 6-\abs{x}\amp=-14\\ 6-\abs{x}\subtractright{6}\amp=-14\subtractright{6}\\ -\abs{x}\amp=-20\\ \multiplyleft{-1}-\abs{x}\amp=\multiplyleft{-1}-20\\ \abs{x}\amp=20\\ x\amp=\pm 20 \end{align*}

The solution set is $\{-20,20\}\text{.}$

###### 11.

Solve $w(y)=10$ where $w(y)=y^2-y-2\text{.}$

Solution
\begin{align*} w(y)\amp=10\\ y^2-y-2\amp=10\\ y^2-y-2\subtractright{10}\amp=10\subtractright{10}\\ y^2-y-12\amp=0\\ (y-4)(y+3)\amp=0 \end{align*}
\begin{align*} y-4\amp=0\amp\amp\text{ or }\amp y+3\amp=0\\ y-4\addright{4}\amp=0\addright{4}\amp\amp\text{ or }\amp y+3\subtractright{3}\amp=0\subtractright{3}\\ y\amp=4\amp\amp\text{ or }\amp y\amp=-3 \end{align*}

The solution set is $\{-3,4\}\text{.}$

###### 12.

Solve $r(t)=s(t)$ where $r(t)=\frac{5}{7}t-3$ and $s(t)=\frac{2}{3}t+\frac{11}{21}\text{.}$

Solution

The solution set is $\{74\}\text{.}$

###### 13.

Solve $g(x)=y(x)$ where $q(x)=6-4x^2$ and $y(x)=(3-x)(8+4x)\text{.}$

Solution

Solve $g(x)=y(x)$ where $q(x)=6-4x^2$ and $y(x)=(3-x)(8+4x)\text{.}$

The solution set is $\{-\frac{9}{2}\}\text{.}$

Determine the function value, or the solution set to the stated equation, based upon the function $f$ shown in Figure 5.3.11.

###### 14.

Determine $f(4)\text{.}$

Solution

$f(4)=2$

###### 15.

Determine $f(2)\text{.}$

Solution

$f(2)=2$

###### 16.

Determine $f(5)\text{.}$

Solution

$f(5)$ is not defined.

###### 17.

Determine $f(1)\text{.}$

Solution

$f(1)=5$

###### 18.

Determine the solution set to $f(x)=-1\text{.}$

Solution

The solution set is $\{-1,3\}\text{.}$

###### 19.

Determine the solution set to $f(x)=2\text{.}$

Solution

The solution set is $\{-4,-2,0,2,4\}\text{.}$

###### 20.

Determine the solution set to $f(x)=-3\text{.}$

Solution

The solution set is $\emptyset\text{.}$

###### 21.

Determine the solution set to $f(x)=5\text{.}$

Solution

The solution set is $\{-3,1\}$

Determine the function value, or the solution set to the stated equation, based upon the function $g$ shown in Figure 5.3.12.

###### 22.

Determine $g(-1)\text{.}$

Solution

$g(-1)=6$

###### 23.

Determine $g(2)\text{.}$

Solution

$g(2)=0$

###### 24.

Determine $g(6)\text{.}$

Solution

$g(6)=-8$

###### 25.

Determine $g(-6)\text{.}$

Solution

$g(-6)$ is not defined.

###### 26.

Determine the solution set to $g(x)=-2\text{.}$

Solution

The solution set is $\{-2,3\}\text{.}$

###### 27.

Determine the solution set to $g(x)=3\text{.}$

Solution

The solution set is $\{\frac{1}{2}\}\text{.}$

###### 28.

Determine the solution set to $g(x)=-6\text{.}$

Solution

The solution set is $\{-4,5\}\text{.}$

###### 29.

Determine the solution set to $g(x)=-10\text{.}$

Solution

The solution set is $\{7\}\text{.}$