6.2: Operaciones sobre funciones dadas por tablas
- Page ID
- 117680
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)También podemos combinar funciones que se definen usando tablas.
Dejar\(f\) y\(g\) ser las funciones definidas por la siguiente tabla.
\ [\ begin {array} {|c||c|c|c|c|c|c|c|c|}
\ hline x & 1 & 2 & 3 & 4 & 5 & 6 & 7\
\ hline\ 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\]
Describa las siguientes funciones a través de una tabla:
- \(2\cdot f(x)+3\)
- \(f(x)-g(x)\)
- \(f(x+2)\)
- \(g(-x)\)
Solución
Para (a) y (b), obtenemos por cálculo inmediato
\ [\ begin {array} {|c||c|c|c|c|c|c|c|c|}
\ hline x & 1 & 2 & 3 & 4 & 5 & 6 & 7
\\ hline\ hline\ hline 2\ cdot f (x) +3 & 15 & 9 & 5 & 11 & 3 & 17 & 15
\\ hline f (x) -g (x) & 2 & 3 & -1 & 2 & 4 & 5\\
\ hline
\ end {array}\ nonumber\]
Por ejemplo, para\(x=3\), obtenemos\(2\cdot f(x)+3=2\cdot f(3)+3=2\cdot 1+3=5\) y\(f(x)-g(x)=f(3)-g(3)=1-2=-1\).
Para la parte (c), tenemos un cálculo similar de\(f(x+2)\). Por ejemplo, para\(x=1\), obtenemos\(f(1+2)=f(1+2)=f(3)=1\).
\ [\ begin {array} {|c||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\]
Obsérvese que para las dos últimas entradas\(x=6\) y\(x=7\) la expresión\(f(x+2)\) es indefinida, ya que, por ejemplo for\(x=6\), es la\(f(x+2)=f(6+2)=f(8)\) que está indefinida. Sin embargo\(x=-1\), para, obtenemos\(f(x+2)=f(-1+2)=f(1)=6\). Si definimos\(h(x)=f(x+2)\), entonces el dominio de\(h\) es por lo tanto\(D_h=\{-1,0,1,2,3,4,5\}\).
Por último, para la parte (d), necesitamos tomar\(x\) como entradas, para lo cual\(g(-x)\) se define vía la tabla para\(g\). Obtenemos la siguiente respuesta.
\ [\ begin {array} {|c||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\]
Dejar\(f\) y\(g\) ser las funciones definidas por la siguiente tabla.
\ [\ begin {array} {|c||c|c|c|c|c|c|}
\ hline x & 1 & 3 & 5 & 7 & 9 & 11\
\ hline\ hline\ hline f (x) & 3 & 5 & 11 & 4 & 9 & 7
\\ hline g (x) & 7 & 6 & 9 & 9 & 9 & 5\
\ hline
\ fin {matriz}\ nonumber\]
Describa las siguientes funciones a través de una tabla:
- \(f\circ g\)
- \(g\circ f\)
- \(f\circ f\)
- \(g\circ g\)
Solución
Las composiciones se calculan mediante evaluación repetida. Por ejemplo,
\[(f\circ g)(1)=f(g(1))=f(7)=4 \nonumber \]
La respuesta completa se muestra a continuación.
\ [\ begin {array} {|c||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 y 9 y 5 &\ texto {undef.} & 9 & amp; 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\]