When attempting to find the inverse of our function, we obtain two \"plus-or-minuses\" in the expression. The correct combination depends on the partition number. For the second \"plus-or-minus\" we have a plus when PartitionNumber = 1 or 4, otherwise we have a minus.
", "name": "SecondSign", "definition": "switch(PartitionNumber = 1 , \"+\" , PartitionNumber = 4 , \"+\" , \"-\")"}, "d": {"group": "Variables for part d and onwards", "templateType": "anything", "description": "A constant that will appear in our function for part d.
", "name": "d", "definition": "random(1..5)"}, "c": {"group": "Variables for part d and onwards", "templateType": "anything", "description": "A constant that will appear in our function for part d.
", "name": "c", "definition": "random(1..5)"}, "FirstSign": {"group": "Variables for parts a to c", "templateType": "anything", "description": "When attempting to find the inverse of our function, we obtain two \"plus-or-minuses\" in the expression. The correct combination depends on the partition number. For the first \"plus-or-minus\" we have a plus when PartitionNumber = 3 or 4, otherwise we have a minus.
", "name": "FirstSign", "definition": "switch(PartitionNumber = 3 , \"+\" , PartitionNumber = 4 , \"+\" , \"-\")"}, "f": {"group": "Variables for part d and onwards", "templateType": "anything", "description": "A constant that will appear in our function for part d.
", "name": "f", "definition": "random(1..5)"}, "RandomStatPoint": {"group": "Variables for parts a to c", "templateType": "anything", "description": "A random stationary point of our function.
", "name": "RandomStatPoint", "definition": "random(a , -1*a , 0)"}, "LocInvCodom": {"group": "Variables for parts a to c", "templateType": "anything", "description": "One of the limits of the codomain of the local inverse will be PolySign*b. LocInvCodom is the other limit. It depends on the value PartitionNumber, since it is PartitionNumber (via LocInvPoint) that tells the student where we want the local inverse to be taken.
", "name": "LocInvCodom", "definition": "switch(PartitionNumber = 1, PolySign*(9*a^4 + b) , PartitionNumber = 2, PolySign*(a^4 + b) , PartitionNumber = 3 , PolySign*(a^4 + b) , PartitionNumber = 4, PolySign*(9*a^4 + b) , '')"}, "b": {"group": "Variables for parts a to c", "templateType": "anything", "description": "This is a constant that is added to our polynomial to ensure that the polynomial does not cross the x-axis. See the question description for the full explanation of our polynomial.
", "name": "b", "definition": "random(1..5)"}, "k": {"group": "Variables for part d and onwards", "templateType": "anything", "description": "A constant that will appear in our function for part e.
", "name": "k", "definition": "random(2..6)"}, "PolySign": {"group": "Variables for parts a to c", "templateType": "anything", "description": "This is a scalar that we will multiply our polynomial by. If PolySign is -1, then the graph of our polynomial will be below the x-axis. If PolySign is 1, then the graph will be above the x-axis. Ultimately, this gives some randomness to the classification of the stationary points of our polynomial. See the question description for the full explanation of our polynomial.
", "name": "PolySign", "definition": "random(-1,1)"}, "LocInvPoint": {"group": "Variables for parts a to c", "templateType": "anything", "description": "This is the point at which we will ask the student to find a local inverse. It is dependent on the PartitionNumber.
", "name": "LocInvPoint", "definition": "switch(PartitionNumber = 1, -a-1 , PartitionNumber = 2, -a/2 , PartitionNumber = 3 , a/2 , PartitionNumber = 4, a+1, '')"}, "LocInvRightDom": {"group": "Variables for parts a to c", "templateType": "anything", "description": "This is the right limit of the of the domain of the local inverse. It depends on the value PartitionNumber, since it is PartitionNumber (via LocInvPoint) that tells the student where we want the local inverse to be taken.
", "name": "LocInvRightDom", "definition": "switch(PartitionNumber = 1, -a , PartitionNumber = 2, 0 , PartitionNumber = 3 , a , PartitionNumber = 4, 2*a, '')"}, "LocInvLeftDom": {"group": "Variables for parts a to c", "templateType": "anything", "description": "This is the left limit of the of the domain of the local inverse. It depends on the value PartitionNumber, since it is PartitionNumber (via LocInvPoint) that tells the student where we want the local inverse to be taken.
", "name": "LocInvLeftDom", "definition": "switch(PartitionNumber = 1, -2*a , PartitionNumber = 2, -a , PartitionNumber = 3 , 0 , PartitionNumber = 4, a, '')"}, "h": {"group": "Variables for part d and onwards", "templateType": "anything", "description": "A constant that will appear in our function for part e.
", "name": "h", "definition": "random(2..6)"}, "g": {"group": "Variables for part d and onwards", "templateType": "anything", "description": "A constant that will appear in our function for part d.
", "name": "g", "definition": "random(2..6)"}, "YSign": {"group": "Variables for parts a to c", "templateType": "anything", "description": "This is the sign that goes in front of y in the local inverse.
", "name": "YSign", "definition": "if(PolySign = -1 , \"-\" , \"\")"}, "j": {"group": "Variables for part d and onwards", "templateType": "anything", "description": "A constant that will appear in our function for part e.
", "name": "j", "definition": "random(2..6)"}, "OppFirstSign": {"group": "Variables for parts a to c", "templateType": "anything", "description": "This is the opposite sign to FirstSign. We need it in one of the questions where we ask students to choose the local inverse from a set of answers.
", "name": "OppFirstSign", "definition": "switch(PartitionNumber = 3 , \"-\" , PartitionNumber = 4 , \"-\" , \"+\")"}, "PartitionNumber": {"group": "Variables for parts a to c", "templateType": "anything", "description": "the domain of our function can be partitioned (more or less) into 4 parts, where the function is invertible when restricted to one of these parts, and each part is as large a possible. PartitionNumber is the parts for which we will require the student to find a local inverse.
\nThe parts are (-\\infty , -a], [-a , 0], [0 , a], [a , \\infty), corresponding to PartitionNumber = 1,2,3,4 respectively.
", "name": "PartitionNumber", "definition": "random(1..4)"}, "OppSecondSign": {"group": "Variables for parts a to c", "templateType": "anything", "description": "This is the opposite sign to SecondSign. We need it in one of the questions where we ask students to choose the local inverse from a set of answers.
", "name": "OppSecondSign", "definition": "switch(PartitionNumber = 1 , \"-\" , PartitionNumber = 4 , \"-\" , \"+\")"}, "a": {"group": "Variables for parts a to c", "templateType": "anything", "description": "The stationary points of our polynomial will be -a,0,a. See question description for the full explanation of our polynomial.
", "name": "a", "definition": "random(2..6)"}}, "variablesTest": {"condition": "", "maxRuns": 100}, "rulesets": {}, "advice": "", "ungrouped_variables": [], "functions": {}, "metadata": {"licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International", "description": "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.
\nPara 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,
"}, "variable_groups": [{"variables": ["a", "b", "PolySign", "RandomStatPoint", "PartitionNumber", "FirstSign", "OppFirstSign", "SecondSign", "OppSecondSign", "LocInvPoint", "LocInvLeftDom", "LocInvRightDom", "LocInvCodom", "YSign"], "name": "Variables for parts a to c"}, {"variables": ["c", "d", "f", "g", "h", "j", "k"], "name": "Variables for part d and onwards"}], "tags": [], "preamble": {"js": "", "css": ""}, "name": "Luis's copy of MA100 MT Week 4", "statement": "Semana 4 (clases 7 y 8): En esta pregunta, verás el boceto del gráfico, los inversos locales y la integración.
\nSi no ha proporcionado una respuesta a cada intervalo de entrada de una pregunta o parte de la pregunta e intenta enviar sus respuestas a la pregunta o la parte, aparecerá el mensaje \"No se puede enviar la respuesta: verifique los errores\". En realidad, su respuesta ha sido enviada, pero al sistema solo le preocupa que no haya enviado una respuesta a cada intervalo de entrada. Por este motivo, asegúrese de proporcionar una respuesta a cada brecha de entrada en la pregunta o parte antes de enviarla. Incluso si no está seguro de la respuesta, escriba lo que cree que es más probable que sea correcto; Siempre puedes cambiar tu respuesta o volver a intentar la pregunta.
\nAl igual que con todas las preguntas, puede haber partes donde puede elegir \"Mostrar pasos\". Esto puede dar una pista, o puede presentar subpartes que le ayudarán a resolver esa parte de la pregunta. Además, recuerde presionar siempre el botón \"Mostrar comentarios\" al final de cada parte. A veces, se proporcionarán comentarios útiles aquí y, a menudo, dependerán de la respuesta correcta y se vincularán a otras partes de la pregunta. Por lo tanto, siempre vuelva a intentar las partes hasta que obtenga las marcas completas, y luego mire la retroalimentación nuevamente.
Tenga en cuenta que para ver los comentarios de una parte en particular de una pregunta, debe proporcionar una respuesta completa (pero no necesariamente correcta) a esa parte. Sin embargo, no se preocupe, ya que puede ver los comentarios y luego enmendar su respuesta en consecuencia.
Además, al igual que con todas las preguntas, la elección de revelar las respuestas solo le mostrará las respuestas que cambian cada vez que se carga la pregunta (es decir, respuestas a preguntas aleatorias); Las respuestas fijas no serán reveladas.
0
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", "{if(PolySign > 0 , \"maximum\" , \"minimum\")}
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", "{if(PolySign > 0 , \"minimum\" , \"maximum\")}
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", "{if(PolySign > 0 , \"maximum\" , \"minimum\")}
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\nHint for part iii: Recall that we may be able to classify a stationary point by evaluating the second derivative of the function at that point. For example, if $f''(s_1) > 0$ then $s_1$ is a minimum point; and if $f''(s_1) < 0$ then $s_1$ is a maximum point.
", "customMarkingAlgorithm": "", "type": "information", "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "showFeedbackIcon": true, "scripts": {}}], "showCorrectAnswer": true, "extendBaseMarkingAlgorithm": true, "sortAnswers": false, "type": "gapfill", "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "prompt": "Considere la función $ f: [\\ simplify {-2 * {a}}, \\ simplify {2 * {a}}] \\ longrightarrow \\ mathbb {R} $ definido por $ f (x) = \\ simplify [all,! noLeadingMinus] {{PolySign} * x ^ 4 -2 * {PolySign} * {a} ^ 2 * x ^ 2 + {PolySign} * {a} ^ 4 + {PolySign} * {b}} $ para todos $ x \\ en [\\ simplificar {-2 * {a}}, \\ simplificar {2 * {a}}] $.
\nPrimero, 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.
\ni) ¿Cuál es el intercepto y de la función?
ii) Tenga en cuenta que podemos ver nuestra función como un polinomio en $ x ^ 2 $. Es decir, un polinomio de la forma $ a (x ^ 2) ^ 2 + b (x ^ 2) + c $ para algunos $ a, b, c \\ in \\ mathbb {R} $. ¿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 $ 2x ^ 2 $, 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 $ s_1 $, $ s_2 $ y $ s_3 $, con $ s_1 <s_2 <s_3 $.
$ s_1 = $
$ s_2 = $
$ s_3 = $
Ahora evaluaremos la segunda derivada en estos puntos estacionarios, para determinar si son puntos máximo, mínimo o de inflexión.
\n$ f '' (s_1) = $
$ f '' (s_2) = $
$ f '' (s_3) = $
Does a local inverse for $f$ exist at the point $x_0 = \\var{{RandomStatPoint}}$.
", "Does a local inverse for $f$ exist at the point $x_1 =$ {random(-a-1 ,-a/2 , a/2 , a+1)}
"], "answers": ["Yes
", "No
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\nHence, we can say that our function has no inverse. However there may be local inverses at given points. (We recommend that you look at section 7.4 of the lecture notes for an explanation of local inverse, before continuing.)
\n[[0]]
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", "$y = \\simplify{{FirstSign}} \\sqrt{\\simplify{{a}^2}\\simplify{{OppSecondSign}}\\sqrt{\\var{{YSign}}y-\\var{{b}}}}$
", "$y = \\simplify{{OppFirstSign}} \\sqrt{\\simplify{{a}^2}\\simplify{{SecondSign}}\\sqrt{\\var{{YSign}}y-\\var{{b}}}}$
", "$y = \\simplify{{OppFirstSign}} \\sqrt{\\simplify{{a}^2}\\simplify{{OppSecondSign}}\\sqrt{\\var{{YSign}}y-\\var{{b}}}}$
"], "minMarks": 0, "showCellAnswerState": true, "variableReplacements": [], "scripts": {}, "unitTests": [], "matrix": ["1", 0, 0, 0], "displayType": "radiogroup", "showCorrectAnswer": true, "extendBaseMarkingAlgorithm": true, "displayColumns": "1", "distractors": ["", "", "", ""], "type": "1_n_2", "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "maxMarks": 0}], "variableReplacements": [], "scripts": {}, "unitTests": [], "showCorrectAnswer": true, "extendBaseMarkingAlgorithm": true, "sortAnswers": false, "type": "gapfill", "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "prompt": "Let $x_2 = \\var{{LocInvPoint}}$. We want to find the local inverse of $f$ at this point. First we need to find $A \\subseteq [\\simplify{-2*{a}} , \\simplify{2*{a}}]$ and $B \\subseteq \\mathbb{R}$ such that $f: A \\longrightarrow B$ is invertible, $x_2$ is an interior point of $A$, and $f(x_2)$ is an interior point of $B$. Furthermore, we would like $A$ and $B$ to be as large as possible. Fill in the following gaps to demonstrate what $A$ and $B$ are.
\n$A = \\big[$[[0]] , [[1]]$\\big]$.
$B = \\big[$[[2]] , [[3]]$\\big]$.
Now, if we let $y = f(x) = \\simplify[all , !noLeadingMinus]{{PolySign}* x^4 -2*{PolySign}* {a}^2 * x^2 +{PolySign}*{a}^4 +{PolySign}*{b}}$ for all $x \\in[\\simplify{-2*{a}} ,\\simplify{2*{a}}]$, and try to obtain $x$ in terms of $y$, we would get $x = \\pm \\sqrt{\\simplify{{a}^2} \\pm \\sqrt{\\var{{YSign}} y-\\var{{b}}}}$. There are two occurences of $\\pm$, meaning there are four possible combinations in total. Only one of these combinations will represent the local inverse of $f$ at $x_2 = \\var{{LocInvPoint}}$. The question is: Which one?
\nThere is more than one way to go about answering this. Here is one suggestion: Choose a point $y$ in the interior of $B$, and evaluate $x = \\pm \\sqrt{\\simplify{{a}^2} \\pm \\sqrt{y-\\var{{b}}}}$ for each possible combination. Only one combination will give you a value which lies in $A$. It is therefore this combination that tells us what the local inverse of $f$ at $x_2 = \\var{{LocInvPoint}}$ is.
\nThe local Inverse of $f$ at $x_2 = \\var{{LocInvPoint}}$ is defined by $f^{-1} : B \\longrightarrow A$, where for each $y \\in B$ we have
[[4]]
The remainder of this question involves integration.
\nConsider te function $g:\\mathbb{R} \\longrightarrow \\mathbb{R}$ defined by $ g(x) = \\var{{c}} e^{\\var{{d}}x}$ for $x \\in (-\\infty , 0)$, and $ g(x) = \\frac{\\var{{f}}}{(x+1)^2} - \\frac{\\var{{g}}}{(x+1)^3}$ for $x \\in [0, \\infty)$.
\nComplete the following gaps to calculate $\\int_{-\\infty}^{\\infty} g(x) \\mathrm{d} x$. (For the first two gaps you will need to calculate the primitive functions (these are described at the start of chapter 8 of the notes), and for the third and fourth gaps you will need to evaluate these functions at the limits.)
\n$\\int_{-\\infty}^{\\infty} g(x) \\mathrm{d} x = \\int_{-\\infty}^{0} \\var{{c}} e^{\\var{{d}}x} \\mathrm{d} x + \\int_{0}^{\\infty} \\frac{\\var{{f}}}{(x+1)^2} - \\frac{\\var{{g}}}{(x+1)^3} \\mathrm{d} x = \\Big[$ [[0]] $\\Big]_{-\\infty}^{0} + \\Big[$ [[1]] $\\Big]_{0}^{\\infty} =$ [[2]] $+$ [[3]] $=$ [[4]].
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\nHint: If $x \\in [0, \\infty)$ then the integral $\\int_{-\\infty}^{x} h(u) \\mathrm{d} u$ ranges over $(-\\infty , 0)$ and $[0,x] \\subseteq [0, \\infty)$, and $h(u)$ is defined differently on these two sets. Hence, we must split the the integral into two parts, $\\int_{-\\infty}^{0} h(u) \\mathrm{d} u + \\int_{0}^{x} h(u) \\mathrm{d} u$, and apply the corresponding definition of $h(u)$ to each one.
\nIf $x \\in (-\\infty , 0)$, then the integral $\\int_{-\\infty}^{x} h(u) \\mathrm{d} u$ ranges over $(-\\infty , x) \\subseteq (-\\infty , 0)$ only, and there is only one definition of $h(u)$ on this set. Hence, there is no need to split the integral in this case.
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\nAlso, define the function $H: \\mathbb{R} \\longrightarrow \\mathbb{R}$ by $H(x) = \\int_{-\\infty}^{x} h(u) \\mathrm{d} u$ for all $x \\in \\mathbb{R}$.
\nFill in the gaps below to find an explicit expression for $H(x)$ when $x \\in (-\\infty , 0)$, and another explicit expression when $x \\in [0 , \\infty)$. For the first, third, and fourth gaps you will need to enter the appropriate expression for $h(u)$; press the \"Show steps\" button below for a hint if you need one. (If you wish to write something like $6 \\cos(u)$ then please write 6*cos(u) and not 6cos(u), as the system may not correctly interpret the latter.)
\nWhen $x \\in (-\\infty , 0)$ we have that
$H(x) = \\int_{-\\infty}^{x} h(u) \\mathrm{d} u = \\int_{-\\infty}^{x}$ [[0]] $\\mathrm{d} u =$ [[1]] .
When $x \\in [0 , \\infty)$ we have that
$H(x) = \\int_{-\\infty}^{x} h(u) \\mathrm{d} u = \\int_{-\\infty}^{0}$ [[2]] $\\mathrm{d} u + \\int_{0}^{x}$ [[3]] $\\mathrm{d} u =$ [[4]] + [[5]] .