This document discusses special functions and inverse functions. It defines even, odd, increasing, and decreasing functions. It provides examples of each type of special function and theorems about sums, products, and compositions of functions. It also introduces injective functions and defines them as functions where different inputs always produce different outputs. The goal is to recognize different types of functions and understand properties of inverse functions and their applications in mathematics and engineering.
En este archivo se muestran las consideraciones preliminares para entender limites, tal como factorización, racionalización y valor absoluto. El tema es iniciado con la definición intuitiva, los diferentes teoremas que se aplican en límites, la indeterminación 0/0 y los diversos ejemplos al respecto
En este archivo se muestran las consideraciones preliminares para entender limites, tal como factorización, racionalización y valor absoluto. El tema es iniciado con la definición intuitiva, los diferentes teoremas que se aplican en límites, la indeterminación 0/0 y los diversos ejemplos al respecto
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Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
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June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...Levi Shapiro
Letter from the Congress of the United States regarding Anti-Semitism sent June 3rd to MIT President Sally Kornbluth, MIT Corp Chair, Mark Gorenberg
Dear Dr. Kornbluth and Mr. Gorenberg,
The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
harassment and intimidation at the Massachusetts Institute of Technology (MIT). Failing to act decisively to ensure a safe learning environment for all students would be a grave dereliction of your responsibilities as President of MIT and Chair of the MIT Corporation.
This Congress will not stand idly by and allow an environment hostile to Jewish students to persist. The House believes that your institution is in violation of Title VI of the Civil Rights Act, and the inability or
unwillingness to rectify this violation through action requires accountability.
Postsecondary education is a unique opportunity for students to learn and have their ideas and beliefs challenged. However, universities receiving hundreds of millions of federal funds annually have denied
students that opportunity and have been hijacked to become venues for the promotion of terrorism, antisemitic harassment and intimidation, unlawful encampments, and in some cases, assaults and riots.
The House of Representatives will not countenance the use of federal funds to indoctrinate students into hateful, antisemitic, anti-American supporters of terrorism. Investigations into campus antisemitism by the Committee on Education and the Workforce and the Committee on Ways and Means have been expanded into a Congress-wide probe across all relevant jurisdictions to address this national crisis. The undersigned Committees will conduct oversight into the use of federal funds at MIT and its learning environment under authorities granted to each Committee.
• The Committee on Education and the Workforce has been investigating your institution since December 7, 2023. The Committee has broad jurisdiction over postsecondary education, including its compliance with Title VI of the Civil Rights Act, campus safety concerns over disruptions to the learning environment, and the awarding of federal student aid under the Higher Education Act.
• The Committee on Oversight and Accountability is investigating the sources of funding and other support flowing to groups espousing pro-Hamas propaganda and engaged in antisemitic harassment and intimidation of students. The Committee on Oversight and Accountability is the principal oversight committee of the US House of Representatives and has broad authority to investigate “any matter” at “any time” under House Rule X.
• The Committee on Ways and Means has been investigating several universities since November 15, 2023, when the Committee held a hearing entitled From Ivory Towers to Dark Corners: Investigating the Nexus Between Antisemitism, Tax-Exempt Universities, and Terror Financing. The Committee followed the hearing with letters to those institutions on January 10, 202
Biological screening of herbal drugs: Introduction and Need for
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2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
3. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
𝑅𝑒𝑐𝑜𝑛𝑜𝑐𝑒𝑟 𝑙𝑎𝑠 𝑓𝑢𝑛𝑐𝑖𝑜𝑛𝑒𝑠
𝑝𝑎𝑟, 𝑖𝑚𝑝𝑎𝑟,
𝑐𝑟𝑒𝑐𝑖𝑒𝑛𝑡𝑒 𝑦 𝑑𝑒𝑐𝑟𝑒𝑐𝑖𝑒𝑛𝑡𝑒
𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑟 𝑙𝑎 𝑓𝑢𝑛𝑐𝑖ó𝑛
𝑖𝑛𝑣𝑒𝑟𝑠𝑎
𝐴𝑝𝑙𝑖𝑐𝑎𝑟 𝑙𝑎𝑠 𝑝𝑟𝑜𝑝𝑖𝑒𝑑𝑎𝑑𝑒𝑠
𝑑𝑒 𝑙𝑎 𝑓𝑢𝑛𝑐𝑖ó𝑛 𝑖𝑛𝑣𝑒𝑟𝑠𝑎
4. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Función inversa
En este tema se definen funciones especiales
y la función inversa. Sus aplicaciones se
encuentra en matemática superior e
ingeniería.
Al enviar un mensaje de voz, una persona
emite una señal analógica el cual capta el
celular A, aplica un codificador (𝑓) y lo
transforma en una señal digital, esta señal
viaja hasta su destino llegando como señal
digital al celular B, éste le aplica un
decodificador (𝑓∗
) y lo transforma en una
señal analógica, para que el receptor pueda
escuchar el mensaje.
𝐸𝑚𝑖𝑠𝑜𝑟
𝑆𝑒ñ𝑎𝑙
𝑎𝑛𝑎𝑙ó𝑔𝑖𝑐𝑎
𝑆𝑒ñ𝑎𝑙
𝑎𝑛𝑎𝑙ó𝑔𝑖𝑐𝑎
𝑆𝑒ñ𝑎𝑙
𝑑𝑖𝑔𝑖𝑡𝑎𝑙
𝐴 𝐵
𝑐𝑜𝑑𝑖𝑓𝑖𝑐𝑎𝑑𝑜𝑟
(𝑓)
𝑑𝑒𝑐𝑜𝑑𝑖𝑓𝑖𝑐𝑎𝑑𝑜𝑟
(𝑓∗
)
𝑅𝑒𝑐𝑒𝑝𝑡𝑜𝑟
5. C R E E M O S E N L A E X I G E N C I A
Funciones especiales
C U R S O D E Á L G E B R A
Función par
Diremos que f es par, solo si:
𝑓(−𝑥) = 𝑓(𝑥) ; ∀𝑥 ∈ 𝐷𝑜𝑚𝑓 ∧ −𝑥 ∈ 𝐷𝑜𝑚𝑓
Ejemplos
Son funciones pares, en todo su dominio, las
siguientes funciones
𝑎) 𝑓 𝑥 = 7 ; 𝑥 ∈ ℝ
𝑏) 𝑔 𝑥 = 𝑥2 ; 𝑥 ∈ ℝ
𝑐) ℎ 𝑥 = 𝑥 ; 𝑥 ∈ ℝ
𝑑) 𝑓 𝑥 = 𝑐𝑜𝑠𝑥 ; 𝑥 ∈ ℝ
Observación
Las gráficas de las funciones pares, son
simétricas respecto al eje Y
𝑋
𝑌
𝑥
−𝑥
(𝑥; 𝑓 𝑥 )
(−𝑥; 𝑓 −𝑥 )
6. C R E E M O S E N L A E X I G E N C I A
Función impar
Diremos que f es impar, solo si:
𝑓 −𝑥 = −𝑓(𝑥) ; ∀𝑥 ∈ 𝐷𝑜𝑚𝑓 ∧ −𝑥 ∈ 𝐷𝑜𝑚𝑓
Ejemplos
Son funciones impares, en todo su dominio, las
siguientes funciones
𝑎) 𝑓 𝑥 = 5𝑥 ; 𝑥 ∈ ℝ
𝑏) 𝑔 𝑥 = 𝑥3 ; 𝑥 ∈ ℝ
𝑐) ℎ 𝑥 = 𝑠𝑔𝑛(𝑥) ; 𝑥 ∈ ℝ
𝑑) 𝑓 𝑥 = 𝑠𝑒𝑛𝑥 ; 𝑥 ∈ ℝ
Observación
Las gráficas de las funciones impares, son
simétricas respecto al origen de coordenadas
C U R S O D E Á L G E B R A
𝑋
𝑌
𝑥
−𝑥
𝑓(𝑥)
𝑓(−𝑥)
(𝑥; 𝑓 𝑥 )
(−𝑥; 𝑓 −𝑥 )
7. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Teoremas
1) La suma de dos funciones pares es par.
Tenemos:
𝑓 𝑥 = 𝑥2
; 𝑔 𝑥 = 𝑥4
son funciones pares
→ (𝑓 + 𝑔) 𝑥 = 𝑓(𝑥) + 𝑔 𝑥 = 𝑥2
+ 𝑥4
Es una función par
2) La suma de dos funciones impares es impar.
Tenemos:
𝑓 𝑥 = 𝑥3 ; 𝑔 𝑥 = 𝑥 son funciones impares
→ (𝑓 + 𝑔) 𝑥 = 𝑓(𝑥) + 𝑔 𝑥 = 𝑥3
+ 𝑥
Es una función impar
3) El producto de dos funciones pares es par.
Tenemos:
𝑓 𝑥 = 𝑥2
; 𝑔 𝑥 = 𝑥4
son funciones pares
→ (𝑓. 𝑔) 𝑥 = 𝑓 𝑥 . 𝑔 𝑥 = 𝑥2
. 𝑥4
Es una función par
= 𝑥6
4) El producto de dos funciones impares es par.
Tenemos:
𝑓 𝑥 = 𝑥3 ; 𝑔 𝑥 = 𝑥 son funciones impares
→ (𝑓. 𝑔) 𝑥 = 𝑓 𝑥 . 𝑔 𝑥 = 𝑥3
. 𝑥
Es una función par
= 𝑥4
8. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Funciones monótonas
Función creciente
Una función f es creciente en 𝑎; 𝑏
si para todo 𝑥1; 𝑥2 ∈ 𝑎; 𝑏
𝑥1 < 𝑥2 se cumple 𝑓 𝑥1 < 𝑓(𝑥2)
Ejemplo
La función 𝑓 𝑥 = 𝑥 es creciente; ∀𝑥 ≥ 0
𝑋
𝑌
𝑥1 𝑥2
𝑓(𝑥1)
𝑓(𝑥2)
𝑥1 < 𝑥2 → 𝑓 𝑥1 < 𝑓(𝑥2)
Observación
En una función f creciente y continua,
se cumple:
𝑋
𝑌
𝑎
𝑓(𝑎)
𝑏
𝑓(𝑏)
𝐷𝑜𝑚𝑓 = 𝑎; 𝑏 𝑅𝑎𝑛𝑓 = 𝑓(𝑎); 𝑓(𝑏)
9. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Función decreciente
Una función f es decreciente en 𝑎; 𝑏
si para todo 𝑥1; 𝑥2 ∈ 𝑎; 𝑏
𝑥1 < 𝑥2 se cumple 𝑓 𝑥1 > 𝑓(𝑥2)
Ejemplo
La función 𝑓 𝑥 = −𝑥 es decreciente; ∀𝑥 ≤ 0
𝑋
𝑌
𝑥1 𝑥2
𝑓(𝑥1)
𝑓(𝑥2)
𝑥1 < 𝑥2 → 𝑓 𝑥1 > 𝑓(𝑥2)
Observación
En una función f decreciente y continua,
se cumple:
𝑋
𝑌
𝑎
𝑓(𝑏)
𝑏
𝑓(𝑎)
𝐷𝑜𝑚𝑓 = 𝑎; 𝑏 𝑅𝑎𝑛𝑓 = 𝑓(𝑏); 𝑓(𝑎)
10. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Teoremas
1) La suma de dos funciones crecientes es creciente
Tenemos:
𝑓 𝑥 = 𝑥; 𝑥 ≥ 0 ; 𝑔 𝑥 = 𝑥 son funciones crecientes
→ (𝑓 + 𝑔) 𝑥 = 𝑓(𝑥) + 𝑔 𝑥 = 𝑥 + 𝑥
Es una función creciente en 𝑥 ≥ 0
2) La suma de dos funciones decrecientes es
decreciente
Tenemos:
𝑓 𝑥 =
1
𝑥
; 𝑥 > 0 ; 𝑔 𝑥 = −𝑥 son funciones decrecientes
→ (𝑓 + 𝑔) 𝑥 = 𝑓(𝑥) + 𝑔 𝑥 =
1
𝑥
− 𝑥
Es una función decreciente en 𝑥 > 0
3) Si 𝑓 y 𝑔 son funciones crecientes entonces 𝑓𝑜𝑔
es creciente
Tenemos:
𝑓 𝑥 = 𝑥 + 2; 𝑥 ≥ 0 ; 𝑔 𝑥 = 𝑥 − 1 son funciones crecientes
→ (𝑓𝑜𝑔) 𝑥 = 𝑓(𝑔(𝑥)) = 𝑔(𝑥) + 2 = 𝑥 − 1 + 2
donde 𝑥 ≥ 1 es una función creciente
4) Si 𝑓 y 𝑔 son funciones decrecientes entonces 𝑓𝑜𝑔
es creciente
Tenemos:
𝑓 𝑥 =
1
𝑥
; 𝑥 > 0 ; 𝑔 𝑥 = −𝑥 son funciones decrecientes
→ (𝑓𝑜𝑔) 𝑥 = 𝑓(𝑔(𝑥)) =
1
𝑔 𝑥
=
1
−𝑥
= −
1
𝑥
donde 𝑥 < 0 es una función creciente
11. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
5) Si 𝑓 es creciente y 𝑔 es decreciente entonces 𝑓𝑜𝑔
es decreciente
Tenemos:
𝑓 𝑥 = 𝑥 + 2 crece ; 𝑔 𝑥 =
1
𝑥
decrece
→ (𝑓𝑜𝑔) 𝑥 = 𝑓(𝑔(𝑥)) = 𝑔(𝑥) + 2 =
1
𝑥
+ 2
donde 𝑥 > 0 , es una función decreciente
6) Si 𝑓 es decreciente y 𝑔 es creciente entonces 𝑓𝑜𝑔
es decreciente
Tenemos:
𝑓 𝑥 = −𝑥 + 2 decrece ; 𝑔 𝑥 = 𝑥3
crece
→ (𝑓𝑜𝑔) 𝑥 = 𝑓(𝑔(𝑥)) = −𝑔(𝑥) + 2 = −𝑥3
+ 2
es una función decreciente
Puedes recordar los
resultados de la composición
de funciones crecientes y
decrecientes, trabajando con
la ley de signos de la
multiplicación.
𝐹𝑢𝑛𝑐𝑖ó𝑛 𝑐𝑟𝑒𝑐𝑖𝑒𝑛𝑡𝑒 → + 𝐹𝑢𝑛𝑐𝑖ó𝑛 𝑑𝑒𝑐𝑟𝑒𝑐𝑖𝑒𝑛𝑡𝑒 → −
𝑓 𝑐𝑟𝑒𝑐𝑖𝑒𝑛𝑡𝑒
+
𝑓 𝑑𝑒𝑐𝑟𝑒𝑐𝑖𝑒𝑛𝑡𝑒
−
𝑔 𝑐𝑟𝑒𝑐𝑖𝑒𝑛𝑡𝑒
+
𝑔 𝑑𝑒𝑐𝑟𝑒𝑐𝑖𝑒𝑛𝑡𝑒
−
+
𝑐𝑟𝑒𝑐𝑖𝑒𝑛𝑡𝑒
−
𝑑𝑒𝑐𝑟𝑒𝑐𝑖𝑒𝑛𝑡𝑒
−
𝑑𝑒𝑐𝑟𝑒𝑐𝑖𝑒𝑛𝑡𝑒
+
𝑐𝑟𝑒𝑐𝑖𝑒𝑛𝑡𝑒
; 𝑥 > 0
𝑓𝑜𝑔
12. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Función inyectiva
Llamado también función univalente o uno a uno.
Una función f es inyectiva si a dos elementos
diferentes en el dominio, le corresponden dos
elementos diferentes en el rango, es decir:
𝑆𝑖 𝑎 ≠ 𝑏 → 𝑓 𝑎 ≠ 𝑓(𝑏) ; ∀ 𝑎; 𝑏 ∈ 𝐷𝑜𝑚𝑓
Ejemplo
𝐴 𝐵
𝑓
1 3
2 5
3 7
4 9
𝑓 es inyectiva
𝐶 𝐷
𝑔
1 3
2 5
3 7
4 9
𝑔 no es inyectiva
En forma equivalente:
Una función es inyectiva, si todo elemento de su rango le
corresponde un solo elemento del dominio.
𝑆𝑖 𝑓 𝑎 = 𝑓(𝑏) → 𝑎 = 𝑏 ; 𝑎; 𝑏 ∈ 𝐷𝑜𝑚𝑓
Ejemplo
Pruebe que 𝑓 𝑥 =
2
𝑥 − 1
; 𝑥 ≠ 1 es inyectiva
Resolución
Sean 𝑎; 𝑏 ∈ 𝐷𝑜𝑚𝑓
Si 𝑓 𝑎 = 𝑓(𝑏) →
2
𝑎 − 1
=
2
𝑏 − 1
→ 𝑎 − 1 = 𝑏 − 1 → 𝑎 = 𝑏
∴ 𝑓 es inyectiva
13. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
¡ 𝐈𝐦𝐩𝐨𝐫𝐭𝐚𝐧𝐭𝐞!
En forma gráfica, se dice que una función
f es inyectiva si toda recta horizontal
(paralela al eje X), corta a la gráfica de la
función a lo más en un punto.
Ejemplo
𝑋
𝑌 𝑓
f es inyectiva
𝑋
𝑌 𝑔
g no es inyectiva
Función sobreyectiva
Llamada también función sobre o suryectiva.
La función 𝑓: 𝐴 → 𝐵 es sobreyectiva cuando
el conjunto de llegada 𝐵 es el rango, es decir
𝑓 es sobreyectiva ↔ 𝑅𝑎𝑛𝑓 = 𝐵
Ejemplo
Sea la función 𝑓: 1; 4 → 5; 11
𝑥 → 2𝑥 + 3
¿es sobreyectiva?
Resolución
Tenemos 𝑓 𝑥 = 2𝑥 + 3 ; 𝑥 ∈ 1; 4 hallemos su rango
→ 1 ≤ 𝑥 ≤ 4 → 2 ≤ 2𝑥 ≤ 8 → 5 ≤ 2𝑥 + 3 ≤ 11
→ 𝑅𝑎𝑛𝑓 = 5; 11
Como el rango es igual al conjunto de llegada
f si es sobreyectiva
∴
14. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
¡ 𝐈𝐦𝐩𝐨𝐫𝐭𝐚𝐧𝐭𝐞!
Si en una función no se
indica el conjunto de
llegada, entonces se
asume que el rango es el
conjunto de llegada por
lo tanto la función es
sobreyectiva.
Ejemplo
Dada la función 𝑓 𝑥 = 4𝑥2
− 3𝑥 + 1 ; 𝑥 ∈ ℝ
Como no se indica el conjunto de llegada
→ 𝑓 es sobreyectiva
Función biyectiva
Una función f es biyectiva si es inyectiva y sobreyectiva a la vez
Ejemplo
𝑥
Si 𝑓: 2; 4 → 7; 𝑚
→ 2𝑥 + 𝑛 es una función biyectiva.
Calcule 𝑚. 𝑛
Resolución
Tenemos: 𝑓 𝑥 = 2𝑥 + 𝑛 ; 𝑥 ∈ 2; 4 ; es biyectiva, su gráfica es
𝑋
𝑌
2
7
4
𝑚
𝑓 2 = 2(2) + 𝑛 = 7 → 𝑛 = 3
𝑓 4 = 2(4) + 𝑛 = 𝑚
ቄ
3
→ 𝑚 = 11
∴ 𝑚. 𝑛 = 33
15. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Función inversa
Sea 𝑓 = 𝑥; 𝑦 / 𝑥 ∈ 𝐷𝑜𝑚𝑓, 𝑦 = 𝑓 𝑥 una función
Se define la función inversa 𝑓∗
(o también 𝑓−1
) como:
𝑓∗
= (𝑦; 𝑥) /𝑥 ∈ 𝐷𝑜𝑚𝑓, 𝑦 = 𝑓 𝑥
Propiedades:
1) 𝐷𝑜𝑚𝑓∗
= 𝑅𝑎𝑛𝑓 ∧ 𝑅𝑎𝑛𝑓∗
= 𝐷𝑜𝑚𝑓
2) 𝑓 𝑥 = 𝑦 → 𝑥 = 𝑓∗
𝑦
3) (𝑓∗
𝑜𝑓)(𝑥) = 𝑥 ; 𝑥 ∈ 𝐷𝑜𝑚𝑓
4) (𝑓𝑜𝑓∗
)(𝑥) = 𝑥 ; 𝑥 ∈ 𝐷𝑜𝑚𝑓∗
5) (𝑓∗
)∗
= 𝑓
6) La gráfica de 𝑓∗
es simétrica a la gráfica de 𝑓 respecto
a la recta 𝑦 = 𝑥
𝑋
𝑌
𝑓
𝑓∗
(𝑥; 𝑦)
(𝑦; 𝑥)
𝑃𝑢𝑛𝑡𝑜𝑠 𝑓𝑖𝑗𝑜𝑠
inyectiva.
16. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Ejemplo
Si 𝑓 = (1; 4); (2; −3) ; (−1; 0) ; (−2; 5) ; (4; 1)
𝑓∗
= (4; 1); (−3; 2) ; (0; −1) ; (5; −2); (1; 4)
Entonces
a) 𝐷𝑜𝑚𝑓 = 1 ; 2; −1 ; −2 ; 4 ; 𝑅𝑎𝑛𝑓= 4 ; −3 ; 0 ; 5 ; 1
b) 𝐷𝑜𝑚𝑓∗
= 4 ; −3 ; 0 ; 5 ; 1 ; 𝑅𝑎𝑛𝑓∗
= 1 ; 2; −1 ; −2 ; 4
𝐷𝑜𝑚𝑓∗
= 𝑅𝑎𝑛𝑓 ; 𝑅𝑎𝑛𝑓∗
= 𝐷𝑜𝑚𝑓
Luego: 𝑓(1) = 4 → 1 = 𝑓∗
(4)
𝑓(−3) = 2 → −3 = 𝑓∗
(2)
𝑓(0) = −1 → 0 = 𝑓∗
(−1)
𝑓(5) = −2 → 5 = 𝑓∗
(−2)
𝑓(1) = 4 → 1 = 𝑓∗
(4)
𝑓(𝑎) = 𝑏 → 𝑎 = 𝑓∗
(𝑏)
Además
𝑓∗
𝑜𝑓 = (1; 1); (2; 2); (−1; −1); (−2; −2) ; (4; 4)
𝑓𝑜𝑓∗
= (4; 4); (−3; −3); (0; 0) ; (5; 5) ; (1; 1)
(𝑓∗
𝑜𝑓)(𝑥) = 𝑥 ; 𝑥 ∈ 𝐷𝑜𝑚𝑓
(𝑓𝑜𝑓∗
)(𝑥) = 𝑥 ; 𝑥 ∈ 𝐷𝑜𝑚𝑓∗
Tenemos:
𝑓∗
= (4; 1); (−3; 2) ; (0; −1) ; (5; −2); (1; 4)
𝑓∗ ∗
= (1; 4); (2; −3) ; (−1; 0) ; (−2; 5) ; (4; 1)
𝑓∗ ∗
= 𝑓
Observación: 𝑓 = 1; 2 ; 3; 4 ; 5; 4 ; 6; 7
es una función no inyectiva, su inversa
𝑓∗
= 2; 1 4; 3 4; 5 7; 6 no es función
Por ello es importante garantizar que la función es
inyectiva para realizar el cálculo de la inversa.
17. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Cálculo de la función inversa (𝒇∗
)
Dado la función 𝑦 = 𝑓 𝑥 ; 𝑥 ∈ 𝐷𝑜𝑚𝑓
Para calcular la función inversa, se tiene:
2) Cálculo del 𝐷𝑜𝑚𝑓∗
Para ello, se calcula el rango de f, porque
𝑅𝑎𝑛𝑓 = 𝐷𝑜𝑚𝑓∗
3) Cálculo del 𝑦 = 𝑓∗
(𝑥)
Para ello:
• Se despeja 𝑥 en función de 𝑦
• Se intercambia 𝑥 con 𝑦
Luego, la función inversa queda definida:
𝑦 = 𝑓∗
(𝑥) ; 𝑥 ∈ 𝐷𝑜𝑚𝑓∗
E𝐣𝐞𝐦𝐩𝐥𝐨 𝟏:
Encuentre la función inversa de: 𝑓 𝑥 = 3𝑥 + 1 ; 𝑥 ∈ 1; 5
1) Demostrar que 𝑓 es inyectiva
Resolución
Graficamos la función, tenemos:
𝑋
𝑌
1
𝑓(1)
5
𝑓(5)
𝑓
1) Como la gráfica es una recta, la
función es inyectiva.
2) Tenemos:
𝑓 1 = 3(1) + 1 = 4
𝑓 5 = 3(5) + 1 = 16
→ 𝑅𝑎𝑛𝑓 = 4; 16 = 𝐷𝑜𝑚𝑓∗
4 =
16 =
3) Cálculo del 𝑦 = 𝑓∗
(𝑥)
𝑓 𝑥 = 3𝑥 + 1
𝑦 = → 𝑥 =
𝑦 − 1
3
• Se despeja 𝑥:
• Se intercambia 𝑥 con 𝑦: 𝑦 =
𝑥 − 1
3
= 𝑓∗
(𝑥)
→ 𝑓∗
(𝑥) =
𝑥 − 1
3
; 𝑥 ∈ 4; 16
18. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
E𝐣𝐞𝐦𝐩𝐥𝐨 𝟐:
Encuentre la función inversa de:
𝑓 𝑥 = 𝑥2
− 2𝑥 ; 𝑥 ∈ −6; 0
Resolución
Graficando la función, tenemos:
𝑓 𝑥 = 𝑥2
− 2𝑥 = (𝑥 − 1)2
− 1 = 𝑥(𝑥 − 2)
𝑋
𝑌
0 2
−6
𝑓 −6
𝑓
1) Por su gráfica, es una
función inyectiva
2) Tenemos:
𝑓 −6 = (−6)2
−2(−6) = 50
𝑓 0 = (0)2
−2(0) = 0
→ 𝑅𝑎𝑛 𝑓 = 0; 50 = 𝐷𝑜𝑚𝑓∗
3) Cálculo del 𝑦 = 𝑓∗
(𝑥)
• Se despeja 𝑥: 𝑓 𝑥 = 𝑥2
− 2𝑥
𝑦 = = (𝑥 − 1)2
+ 1
→ (𝑥 − 1)2
+ 1 = 𝑦 ; 𝑥 ∈ −6; 0
→ (𝑥 − 1)2
= 𝑦 − 1 ; −6 ≤ 𝑥 ≤ 0
→ 𝑥 − 1 = 𝑦 − 1 ∨ 𝑥 − 1 = − 𝑦 − 1
→ 𝑥 = 1 + 𝑦 − 1 ∨ 𝑥 = 1 − 𝑦 − 1
No cumple −6 ≤ 𝑥 ≤ 0
• Se intercambia 𝑥 con 𝑦: 𝑦 = 1 − 𝑥 − 1 = 𝑓∗
(𝑥)
∴ 𝑓∗
(𝑥) = 1 − 𝑥 − 1 ; 𝑥 ∈ 0; 50
19. C R E E M O S E N L A E X I G E N C I A
E𝐣𝐞𝐦𝐩𝐥𝐨 𝟑:
Encuentre la gráfica de la función inversa de:
𝑓 𝑥 = 𝑥3
+ 1
Resolución
En forma gráfica tenemos:
𝑋
𝑌
𝑓(𝑥) = 𝑥3
+ 1
1
𝑦 = 𝑥
1
𝒚 = 𝒇∗
(𝒙)
C U R S O D E Á L G E B R A
Teoremas
Sean 𝑓, 𝑔 funciones biyectivas, se cumple:
1) 𝑘𝑓(𝑥) ∗
=
1
𝑘
. 𝑓∗
(𝑥) ; 𝑘 ∈ ℝ − 0
E𝐣𝐞𝐦𝐩𝐥𝐨 Si 𝑓 𝑥 = 𝑥3
→ 𝑓∗
𝑥 = 3
𝑥
Luego: = 5𝑓(𝑥) ∗
=
1
5
. 𝑓∗
(𝑥) =
1
5
. 3
𝑥
5𝑥3 ∗
2) (𝑓𝑜𝑔)(𝑥) ∗ = (𝑔∗
𝑜𝑓∗
)(𝑥)
E𝐣𝐞𝐦𝐩𝐥𝐨 Si 𝑓 𝑥 = 𝑥3
→ 𝑓∗
𝑥 = 3
𝑥
además 𝑔 𝑥 = 2𝑥 − 1 → 𝑔∗
𝑥 =
𝑥 + 1
2
Luego:
Halle (𝑓𝑜𝑔)(𝑥) ∗
= (𝑔∗
𝑜𝑓∗
)(𝑥) = 𝑔∗
(𝑓∗
(𝑥))
=
𝑓∗
(𝑥) + 1
2
=
3
𝑥 + 1
2
(𝑓𝑜𝑔)(𝑥) ∗
20. w w w . a c a d e m i a c e s a r v a l l e j o . e d u . p e