4.11: Funciones de definición a trozos
- Page ID
- 112500
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Las funciones definidas por partes son funciones que se definen usando diferentes ecuaciones para diferentes partes del dominio.
Evalúe la siguiente función definida por partes para los valores dados de\(x\), y grafique la función:
\(f(x) = \left\{\begin{array}{cc}−2x + 1 & −1 \leq x < 0 \\ x^2 + 2 &0 \leq x \leq 2\end{array} \right.\)
Solución
Para graficar esta función, haz una tabla de soluciones:
|
Tabla de soluciones para\(f(x) = −2x + 1 \) Dominio\(−1 \leq x < 0\) |
|
| \(x\) | \(f(x)\) |
| -1 | 3 |
| 0 | 1 (círculo abierto aquí, 0 no en el dominio) |
|
Tabla de soluciones para\(f(x) = x^2 + 2\) Dominio\(0 \leq x \leq 2\) |
|
| \(x\) | \(f(x)\) |
| 0 | 2 |
| 1 | 3 |
| 2 | 6 |
Evalúe la siguiente función definida por partes para los valores dados de\(x\), y grafique la función:
\(f(x) = \left\{\begin{array}{cc} −x + 1 &x \leq −1 \\ 2 & −1 < x \leq 1 \\ −x + 3 &x > 1 \end{array}\right.\)
Solución
Para graficar esta función, una vez más haga una tabla de soluciones:
|
Tabla de soluciones para\(f(x) = −x + 1\) Dominio\(x \leq −1\) |
|
| \(x\) | \(f(x)\) |
| -3 | 4 |
| -2 | 3 |
| -1 | 2 (círculo cerrado aquí, -1 está en el dominio) |
|
Tabla de soluciones para\(f(x) = 2\) Dominio\(−1 < x \leq 1\) |
|
| \(x\) | \(f(x)\) |
| -1 | 2 (círculo abierto rellenado por la función anterior, -1 no en el dominio) |
| 0 | 2 |
| 1 | 2 (círculo cerrado aquí, 1 está en el dominio) |
|
Tabla de soluciones para\(f(x) = −x + 3\) Dominio\(x > 1\) |
|
| \(x\) | \(f(x)\) |
| 1 | 2 (círculo abierto rellenado por la función anterior, 1 no en el dominio) |
| 2 | 1 |
| 3 | 0 |
Evaluar las siguientes funciones definidas por partes para los valores dados de x, y graficar las funciones:.
- \ (f (x) =\ left\ {\ begin {array} {cc}
x & x<0\\
2 x+1 &x\ geq 0
\ end {array}\ right.\) - \(g(x) = \left\{\begin{array}{cc} 4 − x& x < 2\\ 2x − 2 &x \geq 2 \end{array} \right.\)
- \(h(x) = \left\{\begin{array}{cc} −x − 1 & x < −1 \\ 0& −1 \leq x \leq 1 \\ x + 1 & x > 1 \end{array} \right.\)
- \(g(x) = \left\{\begin{array}{cc} 6 & −8 \leq x < −4 \\ 3 &−4 \leq x \leq 5 \end{array}\right.\)
- \(f(x) = \left\{\begin{array}{cc} −x + 1 & −1 \leq x < 1 \\ \sqrt{x − 1 } &1 \leq x \leq 5\end{array}\right.\)