Inicio › Matemáticas 4 › Representación de funciones

Matemáticas 4 · Tema 2

Representación de funciones

Apuntes

Una misma función se puede ver de varias formas: tabla de valores, parejas ordenadas, diagrama sagital, gráfica en el plano y regla de correspondencia.

Todas dicen lo mismo; pasar de una a otra es representar la función. Cuando el dominio son sólo unos cuantos valores, la recta del plano es una guía: la gráfica son los puntos, por eso va punteada.

Ejemplo

1 · Tabla de valores
xxyy
32
45
58
2 · Parejas ordenadas
S={(3, 2),(4, 5),(5, 8)}S = \{(3,\,2), (4,\,5), (5,\,8)\}
3 · Diagrama sagital
DomRan324558
4 · Gráfica en el plano
5 · Regla de correspondencia
f(x)=3x−7f(x) = \mathbf{3x - 7}
dominio{3,4,5}\{3, 4, 5\}
rango{2,5,8}\{2, 5, 8\}
6 · Comprobación
f(3)=33−7=2f(4)=34−7=5f(5)=35−7=8\begin{aligned} f(3) &= 33 - 7 = 2 \\ f(4) &= 34 - 7 = 5 \\ f(5) &= 35 - 7 = 8 \end{aligned}

Ejercicios

De cada tabla obtén las parejas ordenadas, la regla de correspondencia, el dominio y el rango.

1
xxyy
13
25
-1-1
2
xxyy
1-3
2-5
-11
3
xxyy
1-1
31
42
4
xxyy
1-1
22
35
5
xxyy
1-1
2-3
-13
6
xxyy
11
2-3
3-7
7
xxyy
22
34
-1-4
8
xxyy
-11
-2-1
-3-3
9
xxyy
01
-1-2
14
10
xxyy
0-2
-11
1-5
11
xxyy
0-6
1-2
22
12
xxyy
02
11
3-1

Respuestas

▼
Ejercicio 1
parejas ordenadasS={(1, 3),(2, 5),(−1, −1)}S = \{(1,\,3), (2,\,5), (-1,\,-1)\}
m=5−32−1=2b=3−21=1f(x)=2x+1\begin{aligned} m &= \dfrac{5 - 3}{2 - 1} = 2 \\ b &= 3 - 21 = 1 \\ f(x) &= \mathbf{2x + 1} \end{aligned}
dominio{1,2,−1}\{1, 2, -1\}
rango{3,5,−1}\{3, 5, -1\}
f(1)=21+1=3f(2)=22+1=5f(−1)=2(−1)+1=−1\begin{aligned} f(1) &= 21 + 1 = 3 \\ f(2) &= 22 + 1 = 5 \\ f(-1) &= 2(-1) + 1 = -1 \end{aligned}
Ejercicio 2
parejas ordenadasS={(1, −3),(2, −5),(−1, 1)}S = \{(1,\,-3), (2,\,-5), (-1,\,1)\}
m=−5−(−3)2−1=−2b=−3−(−2)1=−1f(x)=−2x−1\begin{aligned} m &= \dfrac{-5 - (-3)}{2 - 1} = -2 \\ b &= -3 - (-2)1 = -1 \\ f(x) &= \mathbf{-2x - 1} \end{aligned}
dominio{1,2,−1}\{1, 2, -1\}
rango{−3,−5,1}\{-3, -5, 1\}
f(1)=(−2)1−1=−3f(2)=(−2)2−1=−5f(−1)=(−2)(−1)−1=1\begin{aligned} f(1) &= (-2)1 - 1 = -3 \\ f(2) &= (-2)2 - 1 = -5 \\ f(-1) &= (-2)(-1) - 1 = 1 \end{aligned}
Ejercicio 3
parejas ordenadasS={(1, −1),(3, 1),(4, 2)}S = \{(1,\,-1), (3,\,1), (4,\,2)\}
m=1−(−1)3−1=1b=−1−11=−2f(x)=x−2\begin{aligned} m &= \dfrac{1 - (-1)}{3 - 1} = 1 \\ b &= -1 - 11 = -2 \\ f(x) &= \mathbf{x - 2} \end{aligned}
dominio{1,3,4}\{1, 3, 4\}
rango{−1,1,2}\{-1, 1, 2\}
f(1)=11−2=−1f(3)=13−2=1f(4)=14−2=2\begin{aligned} f(1) &= 11 - 2 = -1 \\ f(3) &= 13 - 2 = 1 \\ f(4) &= 14 - 2 = 2 \end{aligned}
Ejercicio 4
parejas ordenadasS={(1, −1),(2, 2),(3, 5)}S = \{(1,\,-1), (2,\,2), (3,\,5)\}
m=2−(−1)2−1=3b=−1−31=−4f(x)=3x−4\begin{aligned} m &= \dfrac{2 - (-1)}{2 - 1} = 3 \\ b &= -1 - 31 = -4 \\ f(x) &= \mathbf{3x - 4} \end{aligned}
dominio{1,2,3}\{1, 2, 3\}
rango{−1,2,5}\{-1, 2, 5\}
f(1)=31−4=−1f(2)=32−4=2f(3)=33−4=5\begin{aligned} f(1) &= 31 - 4 = -1 \\ f(2) &= 32 - 4 = 2 \\ f(3) &= 33 - 4 = 5 \end{aligned}
Ejercicio 5
parejas ordenadasS={(1, −1),(2, −3),(−1, 3)}S = \{(1,\,-1), (2,\,-3), (-1,\,3)\}
m=−3−(−1)2−1=−2b=−1−(−2)1=1f(x)=−2x+1\begin{aligned} m &= \dfrac{-3 - (-1)}{2 - 1} = -2 \\ b &= -1 - (-2)1 = 1 \\ f(x) &= \mathbf{-2x + 1} \end{aligned}
dominio{1,2,−1}\{1, 2, -1\}
rango{−1,−3,3}\{-1, -3, 3\}
f(1)=(−2)1+1=−1f(2)=(−2)2+1=−3f(−1)=(−2)(−1)+1=3\begin{aligned} f(1) &= (-2)1 + 1 = -1 \\ f(2) &= (-2)2 + 1 = -3 \\ f(-1) &= (-2)(-1) + 1 = 3 \end{aligned}
Ejercicio 6
parejas ordenadasS={(1, 1),(2, −3),(3, −7)}S = \{(1,\,1), (2,\,-3), (3,\,-7)\}
m=−3−12−1=−4b=1−(−4)1=5f(x)=−4x+5\begin{aligned} m &= \dfrac{-3 - 1}{2 - 1} = -4 \\ b &= 1 - (-4)1 = 5 \\ f(x) &= \mathbf{-4x + 5} \end{aligned}
dominio{1,2,3}\{1, 2, 3\}
rango{1,−3,−7}\{1, -3, -7\}
f(1)=(−4)1+5=1f(2)=(−4)2+5=−3f(3)=(−4)3+5=−7\begin{aligned} f(1) &= (-4)1 + 5 = 1 \\ f(2) &= (-4)2 + 5 = -3 \\ f(3) &= (-4)3 + 5 = -7 \end{aligned}
Ejercicio 7
parejas ordenadasS={(2, 2),(3, 4),(−1, −4)}S = \{(2,\,2), (3,\,4), (-1,\,-4)\}
m=4−23−2=2b=2−22=−2f(x)=2x−2\begin{aligned} m &= \dfrac{4 - 2}{3 - 2} = 2 \\ b &= 2 - 22 = -2 \\ f(x) &= \mathbf{2x - 2} \end{aligned}
dominio{2,3,−1}\{2, 3, -1\}
rango{2,4,−4}\{2, 4, -4\}
f(2)=22−2=2f(3)=23−2=4f(−1)=2(−1)−2=−4\begin{aligned} f(2) &= 22 - 2 = 2 \\ f(3) &= 23 - 2 = 4 \\ f(-1) &= 2(-1) - 2 = -4 \end{aligned}
Ejercicio 8
parejas ordenadasS={(−1, 1),(−2, −1),(−3, −3)}S = \{(-1,\,1), (-2,\,-1), (-3,\,-3)\}
m=−1−1−2−(−1)=2b=1−2(−1)=3f(x)=2x+3\begin{aligned} m &= \dfrac{-1 - 1}{-2 - (-1)} = 2 \\ b &= 1 - 2(-1) = 3 \\ f(x) &= \mathbf{2x + 3} \end{aligned}
dominio{−1,−2,−3}\{-1, -2, -3\}
rango{1,−1,−3}\{1, -1, -3\}
f(−1)=2(−1)+3=1f(−2)=2(−2)+3=−1f(−3)=2(−3)+3=−3\begin{aligned} f(-1) &= 2(-1) + 3 = 1 \\ f(-2) &= 2(-2) + 3 = -1 \\ f(-3) &= 2(-3) + 3 = -3 \end{aligned}
Ejercicio 9
parejas ordenadasS={(0, 1),(−1, −2),(1, 4)}S = \{(0,\,1), (-1,\,-2), (1,\,4)\}
m=−2−1−1−0=3b=1−30=1f(x)=3x+1\begin{aligned} m &= \dfrac{-2 - 1}{-1 - 0} = 3 \\ b &= 1 - 30 = 1 \\ f(x) &= \mathbf{3x + 1} \end{aligned}
dominio{0,−1,1}\{0, -1, 1\}
rango{1,−2,4}\{1, -2, 4\}
f(0)=30+1=1f(−1)=3(−1)+1=−2f(1)=31+1=4\begin{aligned} f(0) &= 30 + 1 = 1 \\ f(-1) &= 3(-1) + 1 = -2 \\ f(1) &= 31 + 1 = 4 \end{aligned}
Ejercicio 10
parejas ordenadasS={(0, −2),(−1, 1),(1, −5)}S = \{(0,\,-2), (-1,\,1), (1,\,-5)\}
m=1−(−2)−1−0=−3b=−2−(−3)0=−2f(x)=−3x−2\begin{aligned} m &= \dfrac{1 - (-2)}{-1 - 0} = -3 \\ b &= -2 - (-3)0 = -2 \\ f(x) &= \mathbf{-3x - 2} \end{aligned}
dominio{0,−1,1}\{0, -1, 1\}
rango{−2,1,−5}\{-2, 1, -5\}
f(0)=(−3)0−2=−2f(−1)=(−3)(−1)−2=1f(1)=(−3)1−2=−5\begin{aligned} f(0) &= (-3)0 - 2 = -2 \\ f(-1) &= (-3)(-1) - 2 = 1 \\ f(1) &= (-3)1 - 2 = -5 \end{aligned}
Ejercicio 11
parejas ordenadasS={(0, −6),(1, −2),(2, 2)}S = \{(0,\,-6), (1,\,-2), (2,\,2)\}
m=−2−(−6)1−0=4b=−6−40=−6f(x)=4x−6\begin{aligned} m &= \dfrac{-2 - (-6)}{1 - 0} = 4 \\ b &= -6 - 40 = -6 \\ f(x) &= \mathbf{4x - 6} \end{aligned}
dominio{0,1,2}\{0, 1, 2\}
rango{−6,−2,2}\{-6, -2, 2\}
f(0)=40−6=−6f(1)=41−6=−2f(2)=42−6=2\begin{aligned} f(0) &= 40 - 6 = -6 \\ f(1) &= 41 - 6 = -2 \\ f(2) &= 42 - 6 = 2 \end{aligned}
Ejercicio 12
parejas ordenadasS={(0, 2),(1, 1),(3, −1)}S = \{(0,\,2), (1,\,1), (3,\,-1)\}
m=1−21−0=−1b=2−(−1)0=2f(x)=−x+2\begin{aligned} m &= \dfrac{1 - 2}{1 - 0} = -1 \\ b &= 2 - (-1)0 = 2 \\ f(x) &= \mathbf{-x + 2} \end{aligned}
dominio{0,1,3}\{0, 1, 3\}
rango{2,1,−1}\{2, 1, -1\}
f(0)=(−1)0+2=2f(1)=(−1)1+2=1f(3)=(−1)3+2=−1\begin{aligned} f(0) &= (-1)0 + 2 = 2 \\ f(1) &= (-1)1 + 2 = 1 \\ f(3) &= (-1)3 + 2 = -1 \end{aligned}

Ejercicios extra

▼
13
xxyy
1-1
21
-1-5
14
xxyy
04
11
2-2
15
xxyy
-23
05
38
16
xxyy
0-1
13
-1-5
17
xxyy
1-3
3-5
-20
18
xxyy
-1-3
02
17
19
xxyy
06
22
30
20
xxyy
1-2
21
-1-8
21
xxyy
-15
01
2-7
22
xxyy
-31
-15
19
23
xxyy
0-1
1-6
-14
24
xxyy
-1-3
03
19
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