{"id":35260,"date":"2024-12-20T13:00:43","date_gmt":"2024-12-20T13:00:43","guid":{"rendered":"https:\/\/toposuranos.com\/material\/?p=35260"},"modified":"2025-12-11T17:06:28","modified_gmt":"2025-12-11T17:06:28","slug":"teorema-de-weierstrass-dos-valores-extremos","status":"publish","type":"post","link":"https:\/\/toposuranos.com\/material\/pt\/teorema-de-weierstrass-dos-valores-extremos\/","title":{"rendered":"Teorema de Weierstrass dos Valores Extremos"},"content":{"rendered":"<style>\np, ul, ol{\ntext-align: justify;\n}\nh1{\ntext-align:center;\ntext-transform: uppercase;\n}\nh2{\ntext-align:center;\ntext-transform: uppercase;\nfont-size:24pt;\n}\nh3 { \n    text-align: center;\n    text-transform: uppercase;\n    font-size: 24px !important;\n}\n<\/style>\n<h1>Teorema de Weierstrass dos Valores Extremos<\/h1>\n<p style=\"text-align:center;\"><em>Por que em tantos problemas de otimiza\u00e7\u00e3o se d\u00e1 quase por certo que \u201co m\u00e1ximo existe\u201d ou que \u201csempre h\u00e1 um m\u00ednimo\u201d em determinado intervalo, quando na realidade nada obriga que isso ocorra? O <strong>Teorema de Weierstrass<\/strong> \u00e9 a pe\u00e7a que faltava nesse quebra-cabe\u00e7a: ele garante que uma fun\u00e7\u00e3o cont\u00ednua definida em um intervalo fechado e limitado n\u00e3o apenas est\u00e1 limitada, mas efetivamente atinge seus valores extremos. Nesta publica\u00e7\u00e3o revisamos seu enunciado, constru\u00edmos em detalhe uma demonstra\u00e7\u00e3o rigorosa baseada em continuidade pontual, compacidade e o axioma do supremo, e comentamos sua interpreta\u00e7\u00e3o moderna em termos de fun\u00e7\u00f5es cont\u00ednuas sobre conjuntos compactos. A ideia \u00e9 que, ao terminar, voc\u00ea n\u00e3o apenas recorde o teorema como uma frase, mas entenda por que ele \u00e9 verdadeiro e por que aparece repetidamente em an\u00e1lise, em otimiza\u00e7\u00e3o e em modelos aplicados.<\/em><\/p>\n<p style=\"text-align:center;\"><b>Objetivos de aprendizagem<\/b><\/p>\n<ol>\n<li>\n    <strong>Compreender o enunciado do Teorema de Weierstrass.<\/strong><br \/>\n    Identificar com precis\u00e3o as hip\u00f3teses do teorema (fun\u00e7\u00e3o cont\u00ednua em um intervalo fechado e limitado <span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span>) e suas conclus\u00f5es principais: limita\u00e7\u00e3o e exist\u00eancia de valores m\u00e1ximo e m\u00ednimo.\n  <\/li>\n<li>\n    <strong>Interpretar o Teorema de Weierstrass em termos de compacidade.<\/strong><br \/>\n    Formular o resultado em linguagem moderna: as fun\u00e7\u00f5es cont\u00ednuas enviam conjuntos compactos a conjuntos nos quais os valores extremos s\u00e3o atingidos, conectando o caso de <span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span> com o marco geral da an\u00e1lise real.\n  <\/li>\n<li>\n    <strong>Relacionar o Teorema de Weierstrass com problemas de otimiza\u00e7\u00e3o.<\/strong><br \/>\n    Reconhecer o papel do teorema como fundamento te\u00f3rico para a exist\u00eancia de m\u00e1ximos e m\u00ednimos em muitos problemas de otimiza\u00e7\u00e3o em uma vari\u00e1vel, tanto em contextos te\u00f3ricos quanto aplicados.\n  <\/li>\n<\/ol>\n<p style=\"text-align:center;\"><b><u>\u00cdNDICE DE CONTE\u00daDOS<\/u>:<\/b><br \/>\n<a href=\"#1\"><b>Introdu\u00e7\u00e3o<\/b><\/a><br \/>\n<a href=\"#2\"><b>Enunciado do Teorema de Weierstrass<\/b><\/a><br \/>\n<a href=\"#3\">Demonstra\u00e7\u00e3o<\/a><br \/>\n<a href=\"#4\">Passo 1: Continuidade pontual em <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/a><br \/>\n<a href=\"#5\">Passo 2: Recobrimento aberto associado \u00e0 continuidade<\/a><br \/>\n<a href=\"#6\">Passo 3: Compacidade de <span dir=\"ltr\">[a,b]<\/span> e subcobertura finita<\/a><br \/>\n<a href=\"#7\">Passo 4: Constru\u00e7\u00e3o de um <span class=\"katex-eq\" data-katex-display=\"false\">\\delta<\/span> que n\u00e3o depende de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span> (continuidade uniforme)<\/a><br \/>\n<a href=\"#8\">Passo 5: De continuidade uniforme \u00e0 limita\u00e7\u00e3o de <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> em <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/a><br \/>\n<a href=\"#9\">Passo 6: Exist\u00eancia de valores m\u00e1ximo e m\u00ednimo<\/a><br \/>\n<a href=\"#10\"><b>Interpreta\u00e7\u00e3o em termos de compacidade e conclus\u00e3o<\/b><\/a>\n<\/p>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/N5mSrhJgCds\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><br \/>\n<a name=\"1\"><\/a><\/br><\/p>\n<h2>Introdu\u00e7\u00e3o<\/h2>\n<p>\nO <strong>Teorema de Weierstrass dos Valores Extremos<\/strong> \u00e9 um daqueles resultados que, embora costume aparecer nas primeiras unidades de An\u00e1lise Real, na realidade sustenta silenciosamente uma parte enorme da matem\u00e1tica aplicada. Sempre que em f\u00edsica, economia ou estat\u00edstica falamos em \u201cmaximizar\u201d ou \u201cminimizar\u201d uma quantidade sujeita a determinadas restri\u00e7\u00f5es, no fundo estamos utilizando uma ideia muito pr\u00f3xima da que este teorema garante: que uma fun\u00e7\u00e3o cont\u00ednua definida em um intervalo fechado e limitado <strong>n\u00e3o apenas est\u00e1 limitada, mas efetivamente atinge seus valores extremos<\/strong>.\n<\/p>\n<p>\nIntuitivamente pode parecer \u201c\u00f3bvio\u201d que, se desenhamos uma curva cont\u00ednua sobre um segmento <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>, ent\u00e3o deve existir um ponto mais alto e um mais baixo. No entanto, basta fazer pequenas altera\u00e7\u00f5es nas hip\u00f3teses para que essa intui\u00e7\u00e3o falhe de modo estrondoso: se abrimos o intervalo, se a fun\u00e7\u00e3o deixa de ser cont\u00ednua ou se o dom\u00ednio n\u00e3o \u00e9 limitado, os m\u00e1ximos e m\u00ednimos podem simplesmente desaparecer. O Teorema de Weierstrass organiza essa intui\u00e7\u00e3o e nos diz com precis\u00e3o <em>quando<\/em> podemos confiar nela e <em>por que<\/em>.\n<\/p>\n<p>\nDo ponto de vista te\u00f3rico, este teorema \u00e9 o primeiro encontro s\u00e9rio com a ideia de <strong>compacidade<\/strong>: em linguagem moderna, o que ele afirma \u00e9 que uma fun\u00e7\u00e3o cont\u00ednua transforma conjuntos compactos em conjuntos compactos. Do ponto de vista pr\u00e1tico, isso se traduz na exist\u00eancia de solu\u00e7\u00f5es para muitos problemas de otimiza\u00e7\u00e3o em uma dimens\u00e3o, e ser\u00e1 uma pe\u00e7a fundamental para resultados posteriores, como o <b>Teorema do Valor M\u00e9dio<\/b> e, em \u00faltima inst\u00e2ncia, para compreender com calma o Teorema Fundamental do C\u00e1lculo.\n<\/p>\n<p>\nNesta se\u00e7\u00e3o enunciaremos o Teorema de Weierstrass e desenvolveremos em detalhe sua demonstra\u00e7\u00e3o, apoiando-nos na no\u00e7\u00e3o de continuidade em <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong> e no axioma do supremo. A ideia \u00e9 que este texto te sirva como uma refer\u00eancia s\u00f3lida: tanto para estudar o resultado em si, como para retornar a ele sempre que necessites utiliz\u00e1-lo ao demonstrar outros teoremas ou ao justificar rigorosamente a exist\u00eancia de m\u00e1ximos e m\u00ednimos em problemas concretos.\n<\/p>\n<p><a name=\"2\"><\/a><\/br><\/p>\n<h2>Enunciado do Teorema de Weierstrass<\/h2>\n<table>\n<tbody>\n<tr>\n<td style=\"text-align: justify; background-color: #e0e0ff;\">\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=N5mSrhJgCds&amp;t=439s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Toda fun\u00e7\u00e3o <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> definida<\/span><\/strong><\/a> e cont\u00ednua em <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b],<\/span><\/span> \u00e9 limitada e possui valores m\u00ednimo e m\u00e1ximo, <span class=\"katex-eq\" data-katex-display=\"false\">m<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">M<\/span>, tais que, se <span class=\"katex-eq\" data-katex-display=\"false\">x\\in[a,b]<\/span>, ent\u00e3o <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)\\in[m,M]<\/span><\/span>.<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><a name=\"3\"><\/a><\/br><\/p>\n<h3>Demonstra\u00e7\u00e3o<\/h3>\n<p>\nVamos provar que, se <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f:[a,b]\\to\\mathbb{R}<\/span><\/span><\/strong> \u00e9 cont\u00ednua no intervalo fechado e limitado <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>, ent\u00e3o <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> \u00e9 limitada e atinge um valor m\u00e1ximo e um valor m\u00ednimo em <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>. Dividiremos a demonstra\u00e7\u00e3o em duas grandes partes:\n<\/p>\n<ul>\n<li>Primeiro, mostraremos que a continuidade de <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> em <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong> implica que <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> \u00e9 <em>uniformemente cont\u00ednua<\/em>, e a partir disso deduziremos que \u00e9 <strong>limitada<\/strong>.<\/li>\n<li>Depois, utilizando o axioma do supremo, provaremos que <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> atinge seus valores m\u00e1ximo e m\u00ednimo no intervalo.<\/li>\n<\/ul>\n<p><a name=\"4\"><\/a><\/br><\/p>\n<h4><b>Passo 1:<\/b> Continuidade pontual em <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/h4>\n<p>\nPor hip\u00f3tese, <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> \u00e9 cont\u00ednua em cada ponto <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0\\in[a,b]<\/span><\/span><\/strong>. Pela defini\u00e7\u00e3o de continuidade em termos de <span class=\"katex-eq\" data-katex-display=\"false\">\\epsilon<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">\\delta<\/span>, isso significa que:\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n(\\forall x_0\\in[a,b])(\\forall \\epsilon\\gt 0)(\\exists \\delta(x_0)\\gt 0)\n\n\\big(|x-x_0|\\lt\\delta(x_0)\\Rightarrow |f(x)-f(x_0)|\\lt\\epsilon\\big).\n\n<\/span>\n<\/p>\n<p>\nNeste ponto, o n\u00famero <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta(x_0)<\/span><\/span><\/strong> pode depender do ponto <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span><\/strong>. Nosso objetivo imediato ser\u00e1 construir, a partir desses <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta(x_0)<\/span><\/span><\/strong>, um \u00fanico n\u00famero <strong><span class=\"katex-eq\" data-katex-display=\"false\">\\delta<\/span><\/strong> que n\u00e3o dependa de <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span><\/strong> e que funcione simultaneamente para todos os pontos do intervalo.\n<\/p>\n<p><a name=\"5\"><\/a><\/br><\/p>\n<h4><b>Passo 2:<\/b> Recobrimento aberto associado \u00e0 continuidade<\/h4>\n<p>\nFixemos um <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\epsilon\\gt 0<\/span><\/span><\/strong> qualquer. Para cada <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0\\in[a,b]<\/span><\/span><\/strong>, a continuidade de <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> nos permite escolher um n\u00famero <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta(x_0)\\gt 0<\/span><\/span><\/strong> tal que\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|x-x_0|\\lt\\delta(x_0)\\Rightarrow |f(x)-f(x_0)|\\lt\\frac{\\epsilon}{2}.\n\n<\/span>\n<\/p>\n<p>\nA partir desses valores definimos, para cada <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0\\in[a,b]<\/span><\/span><\/strong>, um intervalo aberto\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\nI_{x_0}=\\left(x_0-\\frac{\\delta(x_0)}{2},\\,x_0+\\frac{\\delta(x_0)}{2}\\right).\n\n<\/span>\n<\/p>\n<p>\nCada <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I_{x_0}<\/span><\/span><\/strong> \u00e9 um conjunto aberto em <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}<\/span> e, al\u00e9m disso, a fam\u00edlia\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n\\{I_{x_0}\\}_{x_0\\in[a,b]}\n\n<\/span>\n<\/p>\n<p>\nforma um <strong>recobrimento aberto<\/strong> de <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>. Com efeito, dado um ponto qualquer <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y\\in[a,b]<\/span><\/span><\/strong>, basta tomar <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0=y<\/span><\/span><\/strong>; por constru\u00e7\u00e3o, <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">y\\in I_y<\/span><\/span><\/strong>. Assim, cada ponto do intervalo pertence a pelo menos um dos abertos <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I_{x_0}<\/span><\/span><\/strong>.\n<\/p>\n<p>\nEssa fam\u00edlia de abertos \u00e9, em geral, <strong>infinita<\/strong> (h\u00e1 um para cada <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0\\in[a,b]<\/span><\/span><\/strong>). \u00c9 aqui que entra em jogo a compacidade de <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>.\n<\/p>\n<p><a name=\"6\"><\/a><\/br><\/p>\n<h4><b>Passo 3:<\/b> Compacidade de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span> e sub-recobrimento finito<\/h4>\n<p>\nSabemos, pelo Teorema de Heine\u2013Borel, que um subconjunto de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}<\/span><\/span> \u00e9 compacto se, e somente se, \u00e9 fechado e limitado. O intervalo <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong> \u00e9 fechado e limitado; logo, \u00e9 compacto. Pela defini\u00e7\u00e3o de compacidade, isso significa que:\n<\/p>\n<p>\nDe <strong>todo<\/strong> recobrimento aberto de <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong> (mesmo que contenha infinitos conjuntos), pode-se extrair um <strong>sub-recobrimento finito<\/strong>.\n<\/p>\n<p>\nAplicando essa propriedade ao recobrimento aberto <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\{I_{x_0}\\}_{x_0\\in[a,b]}<\/span><\/span><\/strong>, segue-se que existem pontos <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_1,\\dots,x_N\\in[a,b]<\/span><\/span><\/strong> tais que os intervalos correspondentes\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\nI_{x_1},\\, I_{x_2},\\,\\dots,\\,I_{x_N}\n\n<\/span>\n<\/p>\n<p>\ncontinuam recobrindo todo o intervalo:\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n[a,b]\\subset I_{x_1}\\cup I_{x_2}\\cup\\cdots\\cup I_{x_N}.\n\n<\/span>\n<\/p>\n<p>\nAssim, passamos de uma fam\u00edlia infinita de intervalos abertos para um sub-recobrimento com apenas <strong>um n\u00famero finito<\/strong> de intervalos, sem perder a propriedade de recobrir <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>.\n<\/p>\n<p><a name=\"7\"><\/a><\/br><\/p>\n<h4><b>Passo 4:<\/b> Constru\u00e7\u00e3o de um <span class=\"katex-eq\" data-katex-display=\"false\">\\delta<\/span> que n\u00e3o depende de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span> (continuidade uniforme)<\/h4>\n<p>\nA partir do sub-recobrimento finito definimos o n\u00famero\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n\\delta=\\min\\left\\{\\frac{\\delta(x_1)}{2},\\frac{\\delta(x_2)}{2},\\dots,\\frac{\\delta(x_N)}{2}\\right\\}.\n\n<\/span>\n<\/p>\n<p>\nComo se trata do m\u00ednimo de uma quantidade finita de n\u00fameros positivos, tem-se que <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta\\gt 0<\/span><\/span><\/strong>. Veremos que esse <strong><span class=\"katex-eq\" data-katex-display=\"false\">\\delta<\/span><\/strong> funciona para <strong>todo<\/strong> ponto <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0\\in[a,b]<\/span><\/span><\/strong>, isto \u00e9, n\u00e3o depende da escolha de <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span><\/strong>.\n<\/p>\n<p>\nTomemos agora:\n<\/p>\n<ul>\n<li>um ponto arbitr\u00e1rio <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0\\in[a,b]<\/span><\/span><\/strong>, e<\/li>\n<li>um ponto <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in[a,b]<\/span><\/span><\/strong> tal que <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|x-x_0|\\lt\\delta<\/span><\/span><\/strong>.<\/li>\n<\/ul>\n<p>\nComo os intervalos <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I_{x_1},\\dots,I_{x_N}<\/span><\/span><\/strong> recobrem <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>, o ponto <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span><\/strong> pertence a pelo menos um deles, digamos a <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I_{x_j}<\/span><\/span><\/strong> para algum <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">j\\in\\{1,\\dots,N\\}<\/span><\/span><\/strong>. Pela defini\u00e7\u00e3o de <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I_{x_j}<\/span><\/span><\/strong>, isso significa que\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|x_0-x_j|\\lt\\frac{\\delta(x_j)}{2}.\n\n<\/span>\n<\/p>\n<p>\nAl\u00e9m disso, pela defini\u00e7\u00e3o de <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta<\/span><\/span><\/strong> temos <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta\\le\\frac{\\delta(x_j)}{2}<\/span><\/span><\/strong>, de modo que de <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|x-x_0|\\lt\\delta<\/span><\/span><\/strong> deduzimos\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|x-x_0|\\lt\\frac{\\delta(x_j)}{2}.\n\n<\/span>\n<\/p>\n<p>\nAplicando a desigualdade triangular,\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|x-x_j|\\le |x-x_0|+|x_0-x_j|\n\n\\lt \\frac{\\delta(x_j)}{2}+\\frac{\\delta(x_j)}{2}\n\n=\\delta(x_j).\n\n<\/span>\n<\/p>\n<p>\nPela escolha de <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta(x_j)<\/span><\/span><\/strong> (continuidade de <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> em <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_j<\/span><\/span><\/strong> para o valor <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\epsilon\/2<\/span><\/span><\/strong>), as desigualdades <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|x_0-x_j|\\lt\\delta(x_j)<\/span><\/span><\/strong> e <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|x-x_j|\\lt\\delta(x_j)<\/span><\/span><\/strong> implicam\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|f(x_0)-f(x_j)|\\lt\\frac{\\epsilon}{2}\n\n\\quad\\text{e}\\quad\n\n|f(x)-f(x_j)|\\lt\\frac{\\epsilon}{2}.\n\n<\/span>\n<\/p>\n<p>\nUsando novamente a desigualdade triangular, obtemos\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|f(x)-f(x_0)|\n\n\\le |f(x)-f(x_j)| + |f(x_j)-f(x_0)|\n\n\\lt \\frac{\\epsilon}{2}+\\frac{\\epsilon}{2}\n\n=\\epsilon.\n\n<\/span>\n<\/p>\n<p>\nComo <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span><\/strong> e <strong><span class=\"katex-eq\" data-katex-display=\"false\">x<\/span><\/strong> eram arbitr\u00e1rios, demonstramos que, para o <strong><span class=\"katex-eq\" data-katex-display=\"false\">\\epsilon<\/span><\/strong> fixado no in\u00edcio, existe um <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta\\gt 0<\/span><\/span><\/strong>, independente de <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span><\/strong>, tal que\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n(\\forall x_0\\in[a,b])(\\forall x\\in[a,b])\n\n\\big(|x-x_0|\\lt\\delta\\Rightarrow |f(x)-f(x_0)|\\lt\\epsilon\\big).\n\n<\/span>\n<\/p>\n<p>\nSe renomearmos <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span><\/strong> como <strong><span class=\"katex-eq\" data-katex-display=\"false\">y<\/span><\/strong>, isso se escreve como:\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n(\\forall \\epsilon\\gt 0)(\\exists \\delta\\gt 0)(\\forall x,y\\in[a,b])\n\n\\big(|x-y|\\lt\\delta\\Rightarrow |f(x)-f(y)|\\lt\\epsilon\\big),\n\n<\/span>\n<\/p>\n<p>\nque \u00e9 precisamente a defini\u00e7\u00e3o de <strong>continuidade uniforme<\/strong> de <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> em <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>. No que segue, apenas precisaremos aplicar esse resultado ao caso <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\epsilon=1<\/span><\/span><\/strong>.\n<\/p>\n<p><a name=\"8\"><\/a><\/br><\/p>\n<h4><b>Passo 5:<\/b> De continuidade uniforme \u00e0 limita\u00e7\u00e3o de <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> em <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/h4>\n<p>\nApliquemos agora a continuidade uniforme com <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\epsilon=1<\/span><\/span><\/strong>. Existe um n\u00famero <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta_1\\gt 0<\/span><\/span><\/strong> tal que, para todos <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x,y\\in[a,b]<\/span><\/span><\/strong>, vale\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|x-y|\\lt\\delta_1\\Rightarrow |f(x)-f(y)|\\lt 1.\n\n<\/span>\n<\/p>\n<p>\nDividimos agora o intervalo <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong> em uma quantidade finita de subintervalos cuja extens\u00e3o seja menor que <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\delta_1<\/span><\/span><\/strong>. Ou seja, escolhemos um inteiro <strong><span class=\"katex-eq\" data-katex-display=\"false\">n<\/span><\/strong> e pontos\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\na = x_0 \\lt x_1 \\lt \\cdots \\lt x_n = b\n\n<\/span>\n<\/p>\n<p>\nde modo que, para cada <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=0,1,\\dots,n-1]<\/span><\/span><\/strong>, se cumpra\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\nx_{k+1}-x_k\\lt\\delta_1.\n\n<\/span>\n<\/p>\n<p>\nConsideremos agora o conjunto finito de valores\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n\\{f(x_0),f(x_1),\\dots,f(x_{n-1})\\}.\n\n<\/span>\n<\/p>\n<p>\nPor ser um conjunto finito de n\u00fameros reais, podemos definir sem problema\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\nC = \\max\\{|f(x_k)| \\;|\\; k=0,1,\\dots,n-1\\}.\n\n<\/span>\n<\/p>\n<p>\nMostraremos que <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">C+1<\/span><\/span><\/strong> \u00e9 uma cota superior em valor absoluto para <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> em todo o intervalo <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>. Seja <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in[a,b]<\/span><\/span><\/strong> um ponto arbitr\u00e1rio. Ent\u00e3o existe um \u00edndice <strong><span class=\"katex-eq\" data-katex-display=\"false\">k<\/span><\/strong> tal que <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in[x_k,x_{k+1}]<\/span><\/span><\/strong>. Em particular,\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|x-x_k|\\le x_{k+1}-x_k\\lt\\delta_1.\n\n<\/span>\n<\/p>\n<p>\nPela continuidade uniforme com <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\epsilon=1<\/span><\/span><\/strong>, de <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">|x-x_k|\\lt\\delta_1<\/span><\/span><\/strong> segue-se que\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|f(x)-f(x_k)|\\lt 1.\n\n<\/span>\n<\/p>\n<p>\nUtilizando a desigualdade triangular:\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|f(x)|\\le |f(x)-f(x_k)| + |f(x_k)| \\lt 1 + |f(x_k)| \\le 1 + C.\n\n<\/span>\n<\/p>\n<p>\nComo <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in[a,b]<\/span><\/span><\/strong> foi arbitr\u00e1rio, conclu\u00edmos que\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n|f(x)|\\le C+1 \\quad \\text{para todo } x\\in[a,b],\n\n<\/span>\n<\/p>\n<p>\nou seja, a fun\u00e7\u00e3o <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> \u00e9 <strong>limitada<\/strong> em <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>.\n<\/p>\n<p><a name=\"9\"><\/a><\/br><\/p>\n<h4><b>Passo 6:<\/b> Exist\u00eancia de valores m\u00e1ximo e m\u00ednimo<\/h4>\n<p>\nDefinimos o conjunto de valores que a fun\u00e7\u00e3o assume no intervalo:\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\nH=\\{f(x)\\;|\\;x\\in[a,b]\\}\\subset\\mathbb{R}.\n\n<\/span>\n<\/p>\n<p>\nJ\u00e1 sabemos que <strong><span class=\"katex-eq\" data-katex-display=\"false\">H<\/span><\/strong> \u00e9 n\u00e3o vazio (pois <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong> n\u00e3o \u00e9) e limitado, portanto, pelo axioma do supremo, existem n\u00fameros reais\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\nM=\\sup H,\\qquad m=\\inf H.\n\n<\/span>\n<\/p>\n<p>\nVamos provar que <strong><span class=\"katex-eq\" data-katex-display=\"false\">M<\/span><\/strong> \u00e9 atingido como valor da fun\u00e7\u00e3o, isto \u00e9, que existe <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_1\\in[a,b]<\/span><\/span><\/strong> tal que <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x_1)=M<\/span><\/span><\/strong>. Procederemos por redu\u00e7\u00e3o ao absurdo.\n<\/p>\n<p>\nSuponhamos que <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)<\/span><\/span><\/strong> nunca atinja o valor <strong><span class=\"katex-eq\" data-katex-display=\"false\">M<\/span><\/strong>, isto \u00e9:\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n(\\forall x\\in[a,b])\\big(f(x)\\lt M\\big).\n\n<\/span>\n<\/p>\n<p>\nSob essa suposi\u00e7\u00e3o, a fun\u00e7\u00e3o\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\ng(x)=\\frac{1}{M-f(x)}\n\n<\/span>\n<\/p>\n<p>\nest\u00e1 bem definida e \u00e9 positiva para todo <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in[a,b]<\/span><\/span><\/strong>, pois por hip\u00f3tese <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">M-f(x)\\gt 0<\/span><\/span><\/strong>. Al\u00e9m disso, como <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> \u00e9 cont\u00ednua e <strong><span class=\"katex-eq\" data-katex-display=\"false\">M<\/span><\/strong> \u00e9 constante, <strong><span class=\"katex-eq\" data-katex-display=\"false\">g<\/span><\/strong> tamb\u00e9m \u00e9 cont\u00ednua. Pela primeira parte da demonstra\u00e7\u00e3o, toda fun\u00e7\u00e3o cont\u00ednua em <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong> \u00e9 limitada, portanto existe um n\u00famero <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">N\\gt 0<\/span><\/span><\/strong> tal que\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n(\\forall x\\in[a,b])\\big(g(x)\\le N\\big).\n\n<\/span>\n<\/p>\n<p>\nEm particular, para todo <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\in[a,b]<\/span><\/span><\/strong> verifica-se\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n\\frac{1}{M-f(x)} = g(x)\\le N,\n\n<\/span>\n<\/p>\n<p>\no que equivale a\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\nM-f(x)\\ge \\frac{1}{N}\n\n\\quad\\Rightarrow\\quad\n\nf(x)\\le M-\\frac{1}{N}.\n\n<\/span>\n<\/p>\n<p>\nIsso significa que todos os valores de <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(x)<\/span><\/span><\/strong> em <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong> s\u00e3o menores ou iguais a <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">M-\\frac{1}{N}<\/span><\/span><\/strong>. Em particular, o supremo de <strong><span class=\"katex-eq\" data-katex-display=\"false\">H<\/span><\/strong> satisfaz\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\n\\sup H\\le M-\\frac{1}{N}\\lt M,\n\n<\/span>\n<\/p>\n<p>\no que contradiz a defini\u00e7\u00e3o de <strong><span class=\"katex-eq\" data-katex-display=\"false\">M<\/span><\/strong> como supremo de <strong><span class=\"katex-eq\" data-katex-display=\"false\">H<\/span><\/strong>. Portanto, nossa suposi\u00e7\u00e3o inicial era falsa, e deve existir um ponto <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_1\\in[a,b]<\/span><\/span><\/strong> tal que\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\nf(x_1)=M.\n\n<\/span>\n<\/p>\n<p>\nUm racioc\u00ednio totalmente an\u00e1logo, aplicado ao \u00ednfimo <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m=\\inf H<\/span><\/span><\/strong> (por exemplo, considerando a fun\u00e7\u00e3o <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">h(x)=-f(x)<\/span><\/span><\/strong>), demonstra que existe um ponto <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_2\\in[a,b]<\/span><\/span><\/strong> tal que\n<\/p>\n<p style=\"text-align: center;\" dir=\"ltr\">\n<span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\n\nf(x_2)=m.\n\n<\/span>\n<\/p>\n<p><a name=\"10\"><\/a><\/br><\/p>\n<h2>Interpreta\u00e7\u00e3o em termos de compacidade e conclus\u00e3o<\/h2>\n<p>\nDemonstramos que toda fun\u00e7\u00e3o cont\u00ednua <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f:[a,b]\\to\\mathbb{R}<\/span><\/span><\/strong> \u00e9 limitada e atinge seus valores m\u00e1ximo e m\u00ednimo em <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong>. Na linguagem moderna da an\u00e1lise, isso se interpreta dizendo que, em <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}<\/span><\/span><\/strong>, intervalos fechados e limitados como <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">[a,b]<\/span><\/span><\/strong> s\u00e3o conjuntos compactos e que fun\u00e7\u00f5es cont\u00ednuas enviam conjuntos compactos em conjuntos compactos.\n<\/p>\n<p>\nEm particular, se <strong><span class=\"katex-eq\" data-katex-display=\"false\">I<\/span><\/strong> \u00e9 compacto e <strong><span class=\"katex-eq\" data-katex-display=\"false\">f<\/span><\/strong> \u00e9 cont\u00ednua em <strong><span class=\"katex-eq\" data-katex-display=\"false\">I<\/span><\/strong>, ent\u00e3o a imagem <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(I)<\/span><\/span><\/strong> \u00e9 um subconjunto compacto de <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}<\/span><\/span><\/strong>. Isso garante que <strong><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">f(I)<\/span><\/span><\/strong> \u00e9 limitado e que nele s\u00e3o efetivamente atingidos um valor m\u00e1ximo e um valor m\u00ednimo, que \u00e9 precisamente o conte\u00fado do Teorema de Weierstrass.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Teorema de Weierstrass dos Valores Extremos Por que em tantos problemas de otimiza\u00e7\u00e3o se d\u00e1 quase por certo que \u201co m\u00e1ximo existe\u201d ou que \u201csempre h\u00e1 um m\u00ednimo\u201d em determinado intervalo, quando na realidade nada obriga que isso ocorra? O Teorema de Weierstrass \u00e9 a pe\u00e7a que faltava nesse quebra-cabe\u00e7a: ele garante que uma fun\u00e7\u00e3o [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":35255,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":3,"footnotes":""},"categories":[856,571],"tags":[],"class_list":["post-35260","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-calculo-diferencial-pt","category-matematica-pt"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.7 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Teorema de Weierstrass dos Valores Extremos - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Entenda o Teorema de Weierstrass desde o zero: continuidade, compacidade e demonstra\u00e7\u00e3o passo a passo aplicada a problemas de m\u00e1ximos e m\u00ednimos.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/toposuranos.com\/material\/pt\/teorema-de-weierstrass-dos-valores-extremos\/\" \/>\n<meta property=\"og:locale\" content=\"es_ES\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Teorema de Weierstrass dos Valores Extremos\" \/>\n<meta property=\"og:description\" content=\"Entenda o Teorema de Weierstrass desde o zero: continuidade, compacidade e demonstra\u00e7\u00e3o passo a passo aplicada a problemas de m\u00e1ximos e m\u00ednimos.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/toposuranos.com\/material\/pt\/teorema-de-weierstrass-dos-valores-extremos\/\" \/>\n<meta property=\"og:site_name\" content=\"toposuranos.com\/material\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/groups\/toposuranos\" \/>\n<meta property=\"article:published_time\" content=\"2024-12-20T13:00:43+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2025-12-11T17:06:28+00:00\" \/>\n<meta property=\"og:image\" content=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2021\/06\/Weierstrass-1-1024x683.jpg\" \/>\n<meta name=\"author\" content=\"giorgio.reveco\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:title\" content=\"Teorema de Weierstrass dos Valores Extremos\" \/>\n<meta name=\"twitter:description\" content=\"Entenda o Teorema de Weierstrass desde o zero: continuidade, compacidade e demonstra\u00e7\u00e3o passo a passo aplicada a problemas de m\u00e1ximos e m\u00ednimos.\" \/>\n<meta name=\"twitter:image\" content=\"https:\/\/toposuranos.com\/material\/wp-content\/uploads\/2021\/06\/Weierstrass-1.jpg\" \/>\n<meta name=\"twitter:creator\" content=\"@topuranos\" \/>\n<meta name=\"twitter:site\" content=\"@topuranos\" \/>\n<meta name=\"twitter:label1\" content=\"Escrito por\" \/>\n\t<meta name=\"twitter:data1\" content=\"giorgio.reveco\" \/>\n\t<meta name=\"twitter:label2\" content=\"Tiempo de lectura\" \/>\n\t<meta name=\"twitter:data2\" content=\"9 minutos\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/toposuranos.com\/material\/pt\/teorema-de-weierstrass-dos-valores-extremos\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/toposuranos.com\/material\/pt\/teorema-de-weierstrass-dos-valores-extremos\/\"},\"author\":{\"name\":\"giorgio.reveco\",\"@id\":\"https:\/\/toposuranos.com\/material\/#\/schema\/person\/e15164361c3f9a2a02cf6c234cf7fdc1\"},\"headline\":\"Teorema de Weierstrass dos Valores Extremos\",\"datePublished\":\"2024-12-20T13:00:43+00:00\",\"dateModified\":\"2025-12-11T17:06:28+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/toposuranos.com\/material\/pt\/teorema-de-weierstrass-dos-valores-extremos\/\"},\"wordCount\":2760,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\/\/toposuranos.com\/material\/#organization\"},\"image\":{\"@id\":\"https:\/\/toposuranos.com\/material\/pt\/teorema-de-weierstrass-dos-valores-extremos\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/toposuranos.com\/material\/wp-content\/uploads\/2021\/06\/Weierstrass-1.jpg\",\"articleSection\":[\"C\u00e1lculo Diferencial\",\"Matem\u00e1tica\"],\"inLanguage\":\"es\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"https:\/\/toposuranos.com\/material\/pt\/teorema-de-weierstrass-dos-valores-extremos\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/toposuranos.com\/material\/pt\/teorema-de-weierstrass-dos-valores-extremos\/\",\"url\":\"https:\/\/toposuranos.com\/material\/pt\/teorema-de-weierstrass-dos-valores-extremos\/\",\"name\":\"Teorema de Weierstrass dos Valores Extremos - 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