6.3: Ejercicios
- Page ID
- 117678
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Encuentre\(f+g\),\(f-g\),\(f\cdot g\) para las funciones a continuación. Declarar su dominio.
- \(f(x)=x^2+6x\),\(\quad \quad \) y\(g(x)=3x-5\)
- \(f(x)=x^3+5\),\(\quad \quad \) y\(g(x)=5x^2+7\)
- \(f(x)=3x+7\sqrt{x}\),\(\quad \) y\(g(x)=2x^2+5\sqrt{x}\)
- \(f(x)=\dfrac{1}{x+2}\),\(\quad \) y\(g(x)=\dfrac{5x}{x+2}\)
- \(f(x)=\sqrt{x-3}\),\(\quad \quad \) y\(g(x)=2\sqrt{x-3}\)
- \(f(x)=x^2+2x+5\),\(\quad \quad \) y\(g(x)=3x-6\)
- \(f(x)=x^2+3x\),\(\quad \quad \) y\(g(x)=2x^2+3x+4\)
- Contestar
-
- \((f+g)(x)=x^{2}+9 x-5\)con dominio\(D_{f+g}=\mathbb{R},(f-g)(x)= x^2 + 3x + 5\) con dominio\(D_{f-g}=\mathbb{R},(f \cdot g)(x)=3 x^{3}+13 x^{2}-30 x\) con dominio\(D_{f \cdot g}=\mathbb{R}\)
- \((f+g)(x)=x^{3}+5 x^{2}+12, D_{f+g}=\mathbb{R},(f-g)(x)=x^{3}-5 x^{2}-2, D_{f-g}=\mathbb{R},(f \cdot g)(x)=5 x^{5}+7 x^{3}+25 x^{2}+35, D_{f \cdot g}=\mathbb{R}\)
- \((f+g)(x)=2 x^{2}+3 x+12 \sqrt{x}, D_{f+g}=[0, \infty),(f-g)(x)=-2 x^{2}+3 x+2 \sqrt{x}, D_{f-g}=[0, \infty),(f \cdot g)(x)=6 x^{3}+14 x^{2} \sqrt{x}+15 x \sqrt{x}+35 x, D_{f \cdot g}=[0, \infty)\)
- \((f+g)(x)=\dfrac{5 x+1}{x+2}, D_{f+g}=\mathbb{R}-\{-2\},(f-g)(x)=\dfrac{1-5 x}{x+2}, D_{f-g}=\mathbb{R}-\{-2\},(f \cdot g)(x)=\dfrac{5 x}{(x+2)^{2}}, D_{f \cdot g}=\mathbb{R}-\{-2\}\)
- \((f+g)(x)=3 \sqrt{x-3}, D_{f+g}=[3, \infty),(f-g)(x)=-\sqrt{x-3}, D_{f-g}=[3, \infty),(f \cdot g)(x)=2 \cdot(\sqrt{x-3})^{2}=2 \cdot(x-3), D_{f \cdot g}=[3, \infty)\)
- \((f+g)(x)=x^{2}+5 x-1, D_{f+g}=\mathbb{R},(f-g)(x)=x^{2}-x+11, D_{f-g}=\mathbb{R},(f \cdot g)(x)=3 x^{3}+3 x-30, D_{f \cdot g}=\mathbb{R}\)
- \((f+g)(x)=3 x^{2}+6 x+4, D_{f+g}=\mathbb{R},(f-g)(x)=-x^{2}-4, D_{f-g}=\mathbb{R},(f \cdot g)(x)=2 x^{4}+9 x^{3}+13 x^{2}+12 x, D_{f \cdot g}=\mathbb{R}\)
Encuentre\(\dfrac f g\), y\(\dfrac g f\) para las funciones a continuación. Declarar su dominio.
- \(f(x)=3x+6\),\(\quad \quad \) y\(g(x)=2x-8\)
- \(f(x)=x+2\),\(\quad \quad \) y\(g(x)=x^2-5x+4\)
- \(f(x)=\dfrac{1}{x-5}\),\(\quad \quad \) y\(g(x)=\dfrac{x-2}{x+3}\)
- \(f(x)=\sqrt{x+6}\),\(\quad \quad \) y\(g(x)=2x+5\)
- \(f(x)=x^2+8x-33\),\(\quad \quad \) y\(g(x)=\sqrt{x}\)
- Contestar
-
- \(\left(\dfrac{f}{g}\right)(x)=\dfrac{3 x+6}{2 x-8}\)con dominio\(D_{\frac{f}{g}}=\mathbb{R}-\{4\}\),\(\left(\dfrac{g}{f}\right)(x)=\dfrac{2 x-8}{3 x+6}\) con dominio\(D_{\frac{g}{f}}=\mathbb{R}-\{-2\}\)
- \(\left(\dfrac{f}{g}\right)(x)=\dfrac{x+2}{x^{2}-5 x+4}=\dfrac{x+2}{(x-4)(x-1)}\),\(D_{\frac{f}{g}}=\mathbb{R}-\{1,4\}, \quad\left(\dfrac{g}{f}\right)(x)=\dfrac{x^{2}-5 x+4}{x+2}, D_{\frac{g}{f}}=\mathbb{R}-\{-2\}\)
- \(\left(\dfrac{f}{g}\right)(x)=\dfrac{x+3}{(x-5)(x-2)}, D_{\frac{f}{g}}=\mathbb{R}-\{-3,2,5\},\left(\dfrac{g}{f}\right)(x)=\dfrac{(x-5)(x-2)}{x+3}, D_{\frac{g}{f}}=\mathbb{R}-\{-3,5\}\)
- \(\left(\dfrac{f}{g}\right)(x)=\dfrac{\sqrt{x+6}}{2 x+5}, D_{\frac{f}{g}}=\left[-6,-\dfrac{5}{2}\right) \cup\left(-\dfrac{5}{2}, \infty\right), \left(\dfrac{g}{f}\right)(x)=\dfrac{2 x+5}{\sqrt{x+6}}, D_{\frac{g}{f}}=(-6, \infty)\)
- \(\left(\dfrac{f}{g}\right)(x)=\dfrac{x^{2}+8 x-33}{\sqrt{x}}, D_{\frac{f}{g}}=(0, \infty), \left(\dfrac{g}{f}\right)(x)=\dfrac{\sqrt{x}}{x^{2}+8 x-33}, D_{\frac{g}{f}}=[0,3) \cup(3, \infty)\)
Dejar\(f(x)=2x-3\) y\(g(x)=3x^2+4x\). Encuentra las siguientes composiciones
- \(f(g(2))\)
- \(g(f(2))\)
- \(f(f(5))\)
- \(f(5 g(-3))\)
- \(g(f(2)-2)\)
- \(f(f(3)+g(3))\)
- \(g(f(2+x))\)
- \(f(f(-x))\)
- \(f( f(-3)-3 g(2))\)
- \(f(f(f(2)))\)
- \(f(x+h)\)
- \(g(x+h)\)
- Contestar
-
- \(37\)
- \(7\)
- \(11\)
- \(147\)
- \(-1\)
- \(81\)
- \(12 x^{2}+20 x+7\)
- \(-4 x-9\)
- \(-141\)
- \(-5\)
- \(2 x+2 h-3\)
- \(3 x^{2}+6 x h+3 h^{2}+4 x+4 h\)
Encuentra la composición\((f\circ g)(x)\) para las funciones:
- \(f(x)=3x-5\),\(\quad \quad \) y\(g(x)=2x+3\)
- \(f(x)=x^2+2\),\(\quad \quad \) y\(g(x)=x+3\)
- \(f(x)=x^2-3x+2\),\(\quad \quad \) y\(g(x)=2x+1\)
- \(f(x)=x^2+\sqrt{x+3}\),\(\quad \quad \) y\(g(x)=x^2+2x\)
- \(f(x)=\dfrac{2}{x+4}\),\(\quad \quad \) y\(g(x)=x+h\)
- \(f(x)=x^2+4x+3\),\(\quad \quad \) y\(g(x)=x+h\)
- Contestar
-
- \((f \circ g)(x)=6 x+4\)
- \((f \circ g)(x)=x^{2}+6 x+11\)
- \((f \circ g)(x)=4 x^{2}-2x\)
- \((f \circ g)(x)=x^{4}+4 x^{3}+4 x^{2}+\sqrt{x^{2}+2 x+3}\)
- \((f \circ g)(x)=\dfrac{2}{x+h+4}\)
- \((f \circ g)(x)=x^{2}+2 x h+h^{2}+4 x+4 h+3\)
Encuentra las composiciones
\[(f\circ g)(x),\quad (g\circ f)(x),\quad(f\circ f)(x),\quad(g\circ g)(x) \nonumber \]
para las siguientes funciones:
- \(f(x)=2x+4\),\(\quad \quad \) y\(g(x)=x-5\)
- \(f(x)=x+3\),\(\quad \quad \) y\(g(x)=x^2-2x\)
- \(f(x)=2x^2-x-6\),\(\quad \quad \) y\(g(x)=\sqrt{3x+2}\)
- \(f(x)=\dfrac{1}{x+3}\),\(\quad \quad \) y\(g(x)=\dfrac{1}{x}-3\)
- \(f(x)=(2x-7)^2\),\(\quad \quad \) y\(g(x)=\dfrac{\sqrt{x}+7}{2}\)
- Contestar
-
- \((f \circ g)(x)=2 x-6, (g \circ f)(x)=2 x-1, (f \circ f)(x)=4 x+12, (g \circ g)(x)=x-10\)
- \((f \circ g)(x)=x^{2}-2 x+3, (g \circ f)(x)=x^{2}+4 x+3, (f \circ f)(x)=x+6, (g \circ g)(x)=x^{4}-4 x^{3}+2 x^{2}+4 x\)
- \((f \circ g)(x)=6 x-2-\sqrt{3 x+2}, (g \circ f)(x)=\sqrt{6 x^{2}-3 x-16}, (f \circ f)(x)=8 x^{4}-8 x^{3}-48 x^{2}+25 x+72, (g \circ g)(x)=\sqrt{3 \sqrt{3 x+2}+2}\)
- \((f \circ g)(x)=x, (g \circ f)(x)=x, (f \circ f)(x)=\dfrac{x+3}{3 x+10}, (g \circ g)(x)=\dfrac{10 x-3}{1-3 x}\)
- \((f \circ g)(x)=x,(g \circ f)(x)=x,(f \circ f)(x)=\left(2(2 x-7)^{2}-7\right)^{2}\)o expandido en grados decrecientes:\((f \circ f)(x)=64 x^{4}-896 x^{3}+4592 x^{2}-10192 x+8281,(g \circ g)(x)=\dfrac{\sqrt{\frac{\sqrt{x}+7}{2}}+7}{2}=\dfrac{14+\sqrt{14+2 \sqrt{x}}}{4}\)
Dejar\(f\) y\(g\) ser las funciones definidas por la siguiente tabla. Completa la tabla que figura a continuación.
\ [\ 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) & 4 & 5 & 7 & 0 & -2 & 6 & 4
\\ hline g (x) & 6 & -8 & 5 & 2 & 9 & 11 & 2\\
\ hline f (x) +3 & & & & & & & &\
\ hline 4 g (x) +5 & & & & & & & & & &
\ hline g (x) -2 f (x) & & & & & & & & & & & & & &
\ hline f (x+3) & & & & & & &\
\ hline
\ end {array}\ nonumber\]
- Contestar
-
\ (\ 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) & 4 & 5 & 7 & 0 & -2 & 6 & 4
\\ hline g (x) & 6 & -8 & 5 & 2 & 9 & 11 & 2\\
\ hline f (x) +3 & 7 & 8 & 10 & 3 & 1 & 9 & 7\
\ hline 4 g (x) +5 & 29 & -28 & 25 & 13 & 41 & 49 & 13\
\ hline g (x) -2 f (x) & -2 & -18 & -9 & 2 & 13 & -1 y -6\\
\ hline f (x+3) & 0 & -2 & 6 & 4 &\ text {undef.} &\ text {undef.} &\ text {undef.}\\
\ hline
\ end {array}\ nonumber\)Tenga en cuenta, sin embargo, que la tabla completa para\(y = f(x + 3)\) viene dada por:
\ (\ begin {array} {|c||c|c|c|c|c|c|c|c|}
\ hline x & -2 & -1 & 0 & 1 & 2 & 3 & 4\
\ hline\ hline\ hline f (x+3) & 4 & 5 & 7 & 0 & -2 & 6 & 4\
\ hline
\ end {array}\ nonumber\)
Dejar\(f\) y\(g\) ser las funciones definidas por la siguiente tabla. Completa la tabla componiendo las funciones dadas.
\ [\ begin {array} {|c||c|c|c|c|c|c|}
\ hline x & 1 & 2 & 3 & 4 & 5 & 6\
\ hline\ hline\ hline f (x) & 3 & 1 & 2 & 5 & 6 & 3
\\ hline g (x) & 5 & 2 & 6 & 1 & 2 & 4\
\ hline (g\ circ f) (x) y & & & &\
\ hline (f\ circ g) (x) & & & & & & &\
\ hline (f\ circ f) (x) & & & & & & &\
\ hline (g\ circ g) (x) & & & & &\
\ hline
\\ final {matriz}\ nonumber\]
- Contestar
-
\ (\ begin {array} {|c||c|c|c|c|c|c|}
\ hline x & 1 & 2 & 3 & 4 & 5 & 6\
\ hline\ hline\ hline f (x) & 3 & 1 & 2 & 5 & 6 & 3
\\ hline g (x) & 5 & 2 & 6 & 1 & 2 & 4\
\ hline (g\ circ f) (x) y 6 y 5 y 2 y 2 y 4 y 6\
\ hline (f\ circ g) (x) y 6 y 1 y 3 y 3 y 1 y 5\
\ hline (f\ circ f) (x) y 2 y 3 y 1 y 6 y 3 y 2\
\ hline (g\ circ g) (x) y 2 y 2 y 4 & 5 y 2 y 1\\
\ hline
\ end {array}\ nonumber\)
Dejar\(f\) y\(g\) ser las funciones definidas por la siguiente tabla. Completa la tabla componiendo las funciones dadas.
\ [\ begin {array} {|c||c|c|c|c|c|c|c|c|}
\ hline x & 0 & 2 & 4 & 6 & 8 & 10 & 12\
\ hline\ hline\ hline f (x) & 4 & 8 & 5 & 6 & 12 & -1 & 10\
\ hline g (x) & 10 & 2 & 0 & 0 & 6 & 7 & 2 & 8\\
\ hline (g\ circ f) (x) & & & & & & &\
\ hline (f\ circ g) (x) & & & & & & & &
\\ hline (f\ circ f) (x) & & & & & &\
\ hline (g\ circ g) (x) & & & & & ; &\\
\ hline
\ end {array}\ nonumber\]
- Contestar
-
\ (\ begin {array} {|c||c|c|c|c|c|c|c|c|}
\ hline x & 0 & 2 & 4 & 6 & 8 & 10 & 10 & 12\
\ hline\ hline\ hline f (x) & 4 & 8 & 5 & 6 & 12 & -1 & 10
\\ hline g (x) & 10 & 2 & 0 & 6 & 7 & 2 & 8 \
\ hline (g\ circ f) (x) & 0 & 7 &\ text {undef.} & -6 & 8 &\ text {undef.} & 2\\
\ hline (f\ circ g) (x) & -1 & 8 & 4 &\ text {undef.} &\ text {undef.} & 8 & 12
\\ hline (f\ circ f) (x) y 5 amp; 12 &\ text {undef.} & 6 & 10 &\ text {undef.} & -1\\
\ hline (g\ circ g) (x) & 2 & 2 & 10 &\ text {undef.} &\ text {undef.} & 2 & 7\
\ hline
\ end {array}\ nonumber\)