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6.2: Operations on functions given by tables

Last updated

May 2, 2022
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- 6.1: Operations on functions given by formulas
- 6.3: Exercises

Thomas Tradler and Holly Carley
CUNY New York City College of Technology via New York City College of Technology at CUNY Academic Works

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We can also combine functions that are defined using tables.

Example $\PageIndex{1}$

Let $f$ and $g$ be the functions defined by the following table.

$\begin{array}{|c||c|c|c|c|c|c|c|} \hline x & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \hline f(x) & 6 & 3 & 1 & 4 & 0 & 7 & 6 \\ \hline g(x) & 4 & 0 & 2 & 5 & -2 & 3 & 1 \\ \hline \end{array} \nonumber$

Describe the following functions via a table:

$2\cdot f(x)+3$
$f(x)-g(x)$
$f(x+2)$
$g(-x)$

Solution

For (a) and (b), we obtain by immediate calculation

$\begin{array}{|c||c|c|c|c|c|c|c|} \hline x & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \hline 2 \cdot f(x)+3 & 15 & 9 & 5 & 11 & 3 & 17 & 15 \\ \hline f(x)-g(x) & 2 & 3 & -1 & -1 & 2 & 4 & 5 \\ \hline \end{array} \nonumber$

For example, for $x=3$ , we obtain $2\cdot f(x)+3=2\cdot f(3)+3=2\cdot 1+3=5$ and $f(x)-g(x)=f(3)-g(3)=1-2=-1$ .

For part (c), we have a similar calculation of $f(x+2)$ . For example, for $x=1$ , we get $f(1+2)=f(1+2)=f(3)=1$ .

$\begin{array}{|c||c|c|c|c|c|c|c|c|c|} \hline x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & -1 & 0 \\ \hline \hline f(x+2) & 1 & 4 & 0 & 7 & 6 & \text { undef. } & \text { undef. } & 6 & 3 \\ \hline \end{array} \nonumber$

Note that for the last two inputs $x=6$ and $x=7$ the expression $f(x+2)$ is undefined, since, for example for $x=6$ , it is $f(x+2)=f(6+2)=f(8)$ which is undefined. However, for $x=-1$ , we obtain $f(x+2)=f(-1+2)=f(1)=6$ . If we define $h(x)=f(x+2)$ , then the domain of $h$ is therefore $D_h=\{-1,0,1,2,3,4,5\}$ .

Finally, for part (d), we need to take $x$ as inputs, for which $g(-x)$ is defined via the table for $g$ . We obtain the following answer.

$\begin{array}{|c||c|c|c|c|c|c|c|} \hline x & -1 & -2 & -3 & -4 & -5 & -6 & -7 \\ \hline \hline g(-x) & 4 & 0 & 2 & 5 & -2 & 3 & 1 \\ \hline \end{array} \nonumber$

Example $\PageIndex{2}$

Let $f$ and $g$ be the functions defined by the following table.

$\begin{array}{|c||c|c|c|c|c|c|} \hline x & 1 & 3 & 5 & 7 & 9 & 11 \\ \hline \hline f(x) & 3 & 5 & 11 & 4 & 9 & 7 \\ \hline g(x) & 7 & -6 & 9 & 11 & 9 & 5 \\ \hline \end{array} \nonumber$

Describe the following functions via a table:

$f\circ g$
$g\circ f$
$f\circ f$
$g\circ g$

Solution

The compositions are calculated by repeated evaluation. For example,

$(f\circ g)(1)=f(g(1))=f(7)=4 \nonumber$

The complete answer is displayed below.

$\begin{array}{|c||c|c|c|c|c|c|} \hline x & 1 & 3 & 5 & 7 & 9 & 11 \\ \hline \hline(f \circ g)(x) & 4 & \text { undef. } & 9 & 7 & 9 & 11 \\ \hline(g \circ f)(x) & -6 & 9 & 5 & \text { undef. } & 9 & 11 \\ \hline(f \circ f)(x) & 5 & 11 & 7 & \text { undef. } & 9 & 4 \\ \hline(g \circ g)(x) & 11 & \text { undef. } & 9 & 5 & 9 & 9 \\ \hline \end{array} \nonumber$

Example 6.2.1\PageIndex{1}

Example 6.2.2\PageIndex{2}

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Example $\PageIndex{1}$

Example $\PageIndex{2}$