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6.2: Operations on functions given by tables

  • Page ID
    48983
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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:

    1. \(2\cdot f(x)+3\)
    2. \(f(x)-g(x)\)
    3. \(f(x+2)\)
    4. \(g(-x)\)

    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:

    1. \(f\circ g\)
    2. \(g\circ f\)
    3. \(f\circ f\)
    4. \(g\circ g\)

    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.

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


    This page titled 6.2: Operations on functions given by tables is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Thomas Tradler and Holly Carley (New York City College of Technology at CUNY Academic Works) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.