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Sea el intervalo $I=[\\var{l},\\var{m}]$ y $g: I \\rightarrow I$ una función definida en este intervalo mediante
\n:\\[g(x) = \\simplify{{c}/3*x^3+ {-c*(a+b)}/2*x^2+{c*a*b}*x+{d}}\\]
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", "partialCredit": 0, "strings": ["(", ")"], "showStrings": false}, "answer": "{c} * (x + {-a}) * (x + {-b})", "answerSimplification": "std", "showCorrectAnswer": true}], "showFeedbackIcon": true, "prompt": "Ingrese la primera derivada de $g$, factorizada en un producto de dos factores lineales en la forma $g'(x) = c(x-a)(x-b)$ para enteros adecuados $a$, $b$ y $c$ :
\n\n$g'(x)=\\;\\;$[[0]]
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\nPunto estacionario más pequeño: [[0]] Punto estacionario más grande: [[1]]
\n¿Ambos puntos estacionarios se encuentran en el intervalo $I$ ? [[2]]
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\n$g''(x)=\\;\\;$ [[0]]
\nEncuentre todos los máximos y mínimos locales dados por los puntos estacionarios.
\nEl máximo local está en $x=\\;\\;$ [[1]] y el valor de la función en el máximo local es = [[2]]
\nEl mínimo local está en $x=\\;\\;$ [[3]] y el valor de la función en el mínimo local es = [[4]]
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\n$g(\\var{l})=\\;\\;$ [[0]] $g(\\var{m})=\\;\\;$ [[1]]
\nIngresar ambas con 3 decimales.
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\n¿Qué valor de $x$ en $I$, $g$ tiene un máximo global?
\n$x=\\;\\;$ [[0]]
\nValor de $g$ en este máximo global = [[1]] (ingresar 3 decimales).
\nMínimo global
\n¿Qué valor de $x$ en $I$, $g$ tiene un mínimo global?
\n$x=\\;\\;$ [[2]]
\nValor de $g$ en este mínimo global = [[3]] (ingresar 3 decimales).
", "showCorrectAnswer": true, "extendBaseMarkingAlgorithm": true, "unitTests": [], "scripts": {}, "type": "gapfill", "variableReplacements": []}], "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "$I$ compact interval, $g:I\\rightarrow I,\\;g(x)=ax^3+bx^2+cx+d$. Find stationary points, local and global maxima and minima of $g$ on $I$
\n$ I $ intervalo compacto, $ g: I \\ rightarrow I, \\; g (x) = ax ^ 3 + bx ^ 2 + cx + d $. Encuentre puntos estacionarios, máximos locales y globales y mínimos de $ g $ en $ I $
"}, "advice": "Diferenciando,tenemos
\n\\[g'(x)=\\simplify{{c}*x^2+{-c*(a+b)}*x+{c*a*b}={c}*(x+{-a})(x-{b})}\\]
\nTenga en cuenta que ya hemos factorizado la derivada.
\nLos puntos estacionarios se obtienen resolviendo. $g'(x)=0 \\Rightarrow x=\\var{a},\\;\\;\\mbox {o } x=\\var{b}$
\nEl menor punto estacionario es $x=\\var{a}$ y el más grande es $x=\\var{b}$.
\nPuesto $\\var{a} > \\var{l}$ y $\\var{b} \\lt \\var{m}$ tenemos que ambos puntos estacionarios están en $I$.
\nLa segunda derivada viene dada por \\[g''(x)=\\simplify{{2*c}*x-{c*(a+b)}}\\]
\nEn el punto estacionario $x=\\var{a}$ tenemos $g''(\\var{a})=\\var{c*a-c*b} \\lt 0$.
\npor lo tanto en este valor de $x$ tenemos un máximo local.
\nEl valor de la función $g$ en este máximo local es $g(\\var{a})= \\var{valmax}$.
\nen el punto estacionario $x=\\var{b}$ tenemos $g''(\\var{b})=\\var{c*b-c*a} \\gt 0$.
\nen consecuencia este punto es un mínimo local.
\nEl valor de la función $g$ en este mínimo local es $g(\\var{b})= \\var{valmin}$.
\nPrimero encontramos los valores en los extremos del intervalo. $I=[\\var{l},\\var{m}]$ son:
\n$g(\\var{l})=\\var{valbegin}$ con 3 decimales.
\n$g(\\var{m})=\\var{valend}$ con 3 decimales.
\nPara encontrar el máximo global, tenga en cuenta que solo nos preocupan los valores de $g$ en el intervalo de $I$
\nProcedemos a comparar los valores de la función en los extremos del intervalo con el máximo local.
\na) Si el valor en el máximo local es mayor que cualquiera de los valores en los extremos, entonces este es el máximo global en el intervalo.
b) De lo contrario, si el mayor valor de la función en los puntos extremos es mayor que el máximo local, entonces este es el máximo global.
\n
\\[\\begin{array}{c|c|c|c} x & \\mbox{máximo local}=\\var{a} & \\var{l} \\in I & \\var{m} \\in I \\\\ \\hline\\\\ g(x)& \\var{valmax} & \\var{valbegin} & \\var{valend} \\\\ \\end{array} \\]
\nPara nuestro ejemplo, vemos que el máximo global ocurre en $x=\\var{gma}$ y $g(\\var{gma})=\\var{valgmax}$.
\nProcedemos como en el máximo global, comparando los valores de la función en los extremos del intervalo con el mínimo local.
a) Si el valor en el mínimo local es menor que cualquiera de los valores en los extremos del intervalo, entonces este es el mínimo global en el intervalo.
b) De lo contrario, si el valor mínimo de la función en los extremos del intervalo es menor que el mínimo local, entonces este es el mínimo global.
\\[\\begin{array}{c|c|c|c} x & \\mbox{mínimo local}=\\var{b} & \\var{l} \\in I & \\var{m} \\in I \\\\ \\hline\\\\ g(x)& \\var{valmin} & \\var{valbegin} & \\var{valend} \\\\ \\end{array} \\]
\nPara nuestro ejemplo, vemos que el máximo global ocurre en t $x=\\var{gmi}$ y $g(\\var{gmi})=\\var{valgmin}$.
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