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Section4.2Introduction to Function Notation

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 TableĀ 4.2.1. 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*}
\(x\) \(y\)
\(0\) \(7\)
\(1\) \(-3\)
\(2\) \(11\)
\(3\) \(7\)
Table4.2.1The ordered pairs in \(f\)

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 FigureĀ 4.2.2. 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*}
plain text
Figure4.2.2\(y=g(x)\)

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 \(f(4)=0\text{.}\)

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 FigureĀ 4.2.3 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 \(1\text{.}\)

plain text
Figure4.2.3\(y=g(x)\)

Let's consider the equation \(g(x)=-8\text{.}\) We can't draw the line \(y=-8\) onto TableĀ 4.2.4, 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 TableĀ 4.2.4, 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 \(-9\text{.}\)

\(x\) \(y\)
\(-4\) \(2\)
\(-5\) \(0\)
\(-6\) \(-2\)
\(-7\) \(-4\)
\(-8\) \(-6\)
\(-9\) \(-8\)
Table4.2.4\(y=g(x)\)

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 \(4\text{.}\)

Let's see several more examples.

Example4.2.5

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 \(5\text{.}\)

Example4.2.6

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 \(-9\)

Example4.2.7

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).

Example4.2.8

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".

Subsection4.2.1Exercises

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
\begin{align*} h(t)\amp=100\\ t^2-21\amp=100\\ t^2-21\addright{21}\amp=100\addright{21}\\ t^2\amp=121\\ t\amp=\pm\sqrt{121}\\ t\amp=\pm 11 \end{align*}

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
\begin{align*} r(t)\amp=s(t)\\ \frac{5}{7}t-3\amp=\frac{2}{3}t+\frac{11}{21}\\ \multiplyleft{21}\left(\frac{5}{7}t-3\right)\amp=\multiplyleft{21}\left(\frac{2}{3}t+\frac{11}{21}\right)\\ 15t-63\amp=14t+11\\ 15t-63\addright{63}\amp=14t+11\addright{63}\\ 15t\amp=14t+74\\ 15t\subtractright{14t}\amp=14t+74\subtractright{14t}\\ t\amp=74 \end{align*}

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 \(y(x)=(3-x)(8+4x)\text{.}\)

\begin{align*} g(x)\amp=y(x)\\ 6-4x^2\amp=(3-x)(8+4x)\\ 6-4x^2\amp=24+4x-4x^2\\ 6-4x^2\addright{4x^2}\amp=24+4x-4x^2\addright{4x^2}\\ 6\amp=24+4x\\ 6\subtractright{24}\amp=24+4x\subtractright{24}\\ -18\amp=4x\\ \divideunder{-18}{4}\amp=\divideunder{4x}{4}\\ -\frac{9}{2}\amp=x \end{align*}

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 FigureĀ 4.2.9.

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 \(\{-3,1\}\)

plain text
Figure4.2.9\(y=f(x)\)

Determine the function value, or the solution set to the stated equation, based upon the function \(g\) shown in FigureĀ 4.2.10.

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 \(\{7\}\text{.}\)

plain text
Figure4.2.10\(y=g(x)\)