6.2: Operations on functions given by tables

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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.

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