6.2: Operations on functions given by tables
- Page ID
- 48983
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)We can also combine functions that are defined using tables.
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.
\[\begin{array}{|c||c|c|c|c|c|c|c|}
\hline x & -1 & -2 & -3 & -4 & -5 & -6 & -7 \\
\hline \hline g(-x) & 4 & 0 & 2 & 5 & -2 & 3 & 1 \\
\hline
\end{array} \nonumber \]
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.
\[\begin{array}{|c||c|c|c|c|c|c|}
\hline x & 1 & 3 & 5 & 7 & 9 & 11 \\
\hline \hline(f \circ g)(x) & 4 & \text { undef. } & 9 & 7 & 9 & 11 \\
\hline(g \circ f)(x) & -6 & 9 & 5 & \text { undef. } & 9 & 11 \\
\hline(f \circ f)(x) & 5 & 11 & 7 & \text { undef. } & 9 & 4 \\
\hline(g \circ g)(x) & 11 & \text { undef. } & 9 & 5 & 9 & 9 \\
\hline
\end{array} \nonumber \]