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Section 5.3 Introduction 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 TableĀ 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*}
\(x\) \(y\)
\(0\) \(7\)
\(1\) \(-3\)
\(2\) \(11\)
\(3\) \(7\)
Figure 5.3.1. The 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 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*}
The graph of a function that resembles a downward facing vee. The vertex of the vee is at the point \((-2,6)\text{.}\) The left side of the vee has an arrow on it. The right side of the vee terminates at a solid dot at the point \((5,-1)\text{.}\) The points \((-4,2)\) and \((1,3)\) also lie on the vee.
Figure 5.3.2. \(y=g(x)\)

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 \(f(4)=0\text{.}\) You can use FigureĀ 5.3.3 to explore this idea further.

Figure 5.3.3. Click and drag the point along the curve. Note the correspondence between the coordinates of the point and the function value at the \(x\)-coordinate of the point.

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 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 \(0\text{.}\) 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 \(3\text{.}\)

The graph of a function that resembles a downward facing vee. The vertex of the vee is at the point \((-2,6)\text{.}\) The left side of the vee has an arrow on it. The right side of the vee terminates at a solid dot at the point \((5,-1)\text{.}\) The dotted line \(y=4\) is also graphed and the line intersects the vee at the points \((-3,4)\) and \((0,4)\text{.}\)
Figure 5.3.4. \(y=g(x)\)

Let's consider the equation \(g(x)=-8\text{.}\) We can't draw the line \(y=-8\) onto TableĀ 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 TableĀ 5.3.5, and we can see that the relevant ordered pair is \((-9,-8)\text{.}\) From this we know that the solution to the equation \(g(x)=-8\) is \(-9\text{.}\)

\(x\) \(y\)
\(-4\) \(2\)
\(-5\) \(0\)
\(-6\) \(-2\)
\(-7\) \(-4\)
\(-8\) \(-6\)
\(-9\) \(-8\)
Figure 5.3.5. \(y=g(x)\)
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 FigureĀ 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.

Figure 5.3.6. Use the slider to change the value of \(x\text{.}\)

Let's see several more examples.

Example 5.3.7.

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 \(5\text{.}\)

Example 5.3.8.

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 \(-9\)

Example 5.3.9.

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).

Example 5.3.10.

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".

Exercises Exercises

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 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 \(\{-3,1\}\)

The graph of a function named f. The function is rather squiggly. The left most point on the function is a solid dot at the point \((-4,2)\text{.}\) The right most point plotted is an open circle at the point \((5,5)\text{.}\) From left to right, the curve starts at the point \((-4,2)\) and rises to a peak at \((-3,5)\text{.}\) It then falls past the point \((-2,2)\) until it reaches a floor at the point \((-1,-1)\text{.}\) It then rises past the point \((0,2)\) until reaching a second peak at the point \((1,5)\text{.}\) The curve then falls past the point \((2,2)\) to a second floor at the point \((3,-1)\text{.}\) The curve finishes up by rising past the point \((4,1)\) before terminating at the hole plotted at the point \((5,5)\text{.}\)
Figure 5.3.11. \(y=f(x)\)

Determine the function value, or the solution set to the stated equation, based upon the function \(g\) shown in FigureĀ 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 \(\{7\}\text{.}\)

The graph of a function named g. There are two distinct pieces to the function. The piece on the left resembles a fishhook. It has open holes at either end. The endpoints are \((-6,2)\) and \((-1,3)\text{.}\) The bottom point on the hook is \((4,-6)\text{.}\) The hook also passes through the point \((-2,-2)\text{.}\) The other piece of the function is a half-line. The half-line originates at a solid dot plotted at the point \((-1,6)\text{.}\) The line then decreases with a slope of \(-2\) passes through the points \((0,4)\text{,}\) \((1,2)\text{,}\) \((2,0)\text{,}\) \((3,-2)\text{,}\) \((4,-4)\) and \((5,-6)\text{.}\) There is an arrow at the right end of the half-line.
Figure 5.3.12. \(y=g(x)\)
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