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Esta es la pregunta para la semana 4 del curso MA100 en el LSE. Examina el material de los capítulos 7 y 8. A continuación se describe cómo se definió un polinomio en la pregunta. Esto puede ser útil para cualquier persona que necesite editar esta pregunta.
Para las partes a a c, utilizamos un polinomio definido como m * (x ^ 4 - 2a ^ 2 x ^ 2 + a ^ 4 + b), donde las variables "a" y "b" se seleccionan al azar de un conjunto de tamaño reajustable, y la variable m se elige aleatoriamente del conjunto {+1, -1}. Podemos ver fácilmente que este polinomio tiene puntos estacionarios en -a, 0 y a. Introdujimos la variable "m" para que estos puntos estacionarios no siempre tuvieran la misma clasificación. La variable "b" es siempre positiva, y esto asegura que nuestro polinomio no cruce el eje x. Los primeros y segundos derivados; puntos estacionarios; la evaluación de la segunda derivada en los puntos estacionarios; la clasificación de los puntos estacionarios; y las intersecciones de los ejes se pueden expresar fácilmente en términos de las variables "a", "b" y "m". En efecto,
Metadata
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England schools
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England university
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Scotland schools
Taxonomy: mathcentre
Taxonomy: Kind of activity
Taxonomy: Context
Contributors
History
Luis Hernandez 6 years, 4 months ago
Created this as a copy of MA100 MT Week 4.| Name | Status | Author | Last Modified | |
|---|---|---|---|---|
| MA100 MT Week 4 | Ready to use | Department of Mathematics LSE | 05/10/2018 23:46 | |
| Luis's copy of MA100 MT Week 4 | draft | Luis Hernandez | 20/12/2018 12:07 |
There is one other version that you do not have access to.
| Name | Type | Generated Value |
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| a | integer |
6
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| b | integer |
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| PolySign | integer |
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| RandomStatPoint | integer |
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| PartitionNumber | integer |
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| FirstSign | string |
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| OppFirstSign | string |
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| SecondSign | string |
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| OppSecondSign | string |
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| LocInvPoint | integer |
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| LocInvLeftDom | integer |
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| LocInvRightDom | integer |
12
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| LocInvCodom | number |
11668
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| YSign | string |
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| Name | Type | Generated Value |
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| c | integer |
3
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| d | integer |
4
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| f | integer |
1
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| g | integer |
3
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| h | integer |
5
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| j | integer |
3
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| k | integer |
4
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| Name | Type | Generated Value |
|---|
Generated value: integer
- LocInvCodom
- LocInvLeftDom
- LocInvPoint
- LocInvRightDom
- RandomStatPoint
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Considere la función f:[ simplify−2∗a, simplify2∗a] longrightarrow mathbbR definido por f(x)= simplify[all,!noLeadingMinus]PolySign∗x4−2∗PolySign∗a2∗x2+PolySign∗a4+PolySign∗b para todos x en[ simplificar−2∗a, simplificar2∗a].
Primero, nos gustaría encontrar las intersecciones de los ejes de esta función, y determinar y clasificar los puntos estacionarios. Esto hará que sea más fácil para usted dibujar el gráfico de la función, lo que ayudará en las partes posteriores de esta pregunta.
i) ¿Cuál es el intercepto y de la función?
ii) Tenga en cuenta que podemos ver nuestra función como un polinomio en x2. Es decir, un polinomio de la forma a(x2)2+b(x2)+c para algunos a,b,c in mathbbR. ¿Cuál es el discriminante de este polinomio?
Por lo tanto, podemos ver que este polinomio (es decir, nuestra función) intercepta el eje x
iii) Ahora encontraremos y clasificaremos los puntos estacionarios. Primero encuentre el primer derivado deivativo y segundo de f.
(Si desea escribir algo así como 2x2, escriba 2 * x ^ 2 y no 2x ^ 2, ya que el sistema puede no interpretarlo correctamente).
F′(x)=
f″(x)=
Ahora podemos usar f′(x) para encontrar los puntos estacionarios. Para esta función hay tres puntos estacionarios que llamaremos s1, s2 y s3, con s1<s2<s3.
s1=
s2=
s3=
Ahora evaluaremos la segunda derivada en estos puntos estacionarios, para determinar si son puntos máximo, mínimo o de inflexión.
f″(s1)=
f″(s2)=
f″(s3)=
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