{"id":118819,"date":"2018-12-21T18:25:19","date_gmt":"2018-12-21T20:25:19","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=118819"},"modified":"2024-06-07T20:15:18","modified_gmt":"2024-06-07T23:15:18","slug":"problema-para-ajudar-na-escola-funcao-par-funcao-impar","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-para-ajudar-na-escola-funcao-par-funcao-impar\/","title":{"rendered":".Problema para ajudar na escola: Fun\u00e7\u00e3o par, fun\u00e7\u00e3o \u00edmpar"},"content":{"rendered":"<p class=\"g2\" ><strong><a href=\"http:\/\/clubes.obmep.org.br\/blog\/?p=220026\" rel=\"noopener\" target=\"_blank\">&#9999; <span style=\"color:#915F6D;font-size:14px;\">Link do problema para dispositivos da <em>Apple<\/em>.<\/span><\/a><\/strong><\/p>\n<div class=\"g1n\" style=\"line-height:27px;\">\n<p class=\"tit\" style=\"line-height:25px;margin:10px\">Problema<br \/>\n<span style=\"font-size:12px;color:#0000FF\"> (A partir da 2\u00aa s\u00e9rie do E. M.  &#8211; <span style=\"font-weight: normal;\"> N\u00edvel de dificuldade: M\u00e9dio<\/span>)<\/span><\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nUma fun\u00e7\u00e3o [tex]f: A \\rightarrow   B[\/tex], com [tex]A,B \\subset \\mathbb{R} [\/tex], \u00e9 dita  uma fun\u00e7\u00e3o par, se[tex] f(-x)=f(x)[\/tex], para todo [tex]x \\in A.[\/tex]<br \/>\nSe [tex] f(-x)=-f(x)[\/tex], para todo [tex] x \\in A[\/tex], ent\u00e3o [tex]f[\/tex] \u00e9 dita uma fun\u00e7\u00e3o \u00edmpar.<\/p>\n<ul>\n<li>Por exemplo, a fun\u00e7\u00e3o <\/li>\n<p>[tex]\\qquad \\begin{align*} f:  \\, &#038; \\mathbb{R} \\rightarrow \\mathbb{R}\\\\ &#038; f(x)=x^2 \\end{align*}[\/tex]<br \/>\n\u00e9 uma fun\u00e7\u00e3o par, j\u00e1 que<br \/>\n[tex]\\qquad f(-x)=(-x)^2=x^2=f(x), \\forall \\, x \\, \\in \\, \\mathbb{R}.[\/tex]<\/p>\n<li>J\u00e1 a fun\u00e7\u00e3o <\/li>\n<p>[tex]\\qquad \\begin{align*} f:  \\, &#038; \\mathbb{R} \\rightarrow \\mathbb{R}\\\\ &#038; f(x)=x^3 \\end{align*}[\/tex]<br \/>\n\u00e9 \u00edmpar, uma vez que<br \/>\n[tex]\\qquad f(-x)=(-x)^3=\\left((-1) \\cdot (x)\\right)^3=(-1)^3 \\cdot (x)^3=-\\left(x^3\\right)=-f(x), \\forall \\, x \\, \\in \\, \\mathbb{R}.[\/tex]<\/p>\n<li>No entanto a fun\u00e7\u00e3o <\/li>\n<p>[tex]\\qquad \\begin{align*} f:  \\, &#038; \\mathbb{R} \\rightarrow \\mathbb{R}\\\\ &#038; f(x)=x^3-1 \\end{align*}[\/tex]<br \/>\nn\u00e3o \u00e9 par e nem \u00edmpar. Por exemplo,<br \/>\n[tex]f(-1)=-2 \\ne 0=f(1) \\,  [\/tex] e [tex] \\, f(-1)=-2 \\ne 0=-f(1).  [\/tex]<\/ul>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 30px; margin-bottom: 8px;border: 0 none\"\/>\n<hr style=\"height:6px; width:100%;background-color:#800000; margin-top: 6px; margin-bottom: 8px; border: 0 none\"\/>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 8px; margin-bottom: 30px;border: 0 none\"\/>\nConsidere a fun\u00e7\u00e3o real definida por [tex] f(x)=\\dfrac{x}{1-2^x}-\\dfrac{x}{2}.[\/tex]<br \/>\n<strong>(a)<\/strong> Determine o dom\u00ednio m\u00e1ximo  de [tex] f[\/tex].<br \/>\n<strong>(b)<\/strong> Com esse dom\u00ednio, [tex]f[\/tex] \u00e9 uma fun\u00e7\u00e3o par, ou \u00edmpar ou nem par e nem \u00edmpar?<\/p>\n<p style=\"margin:10px\">\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1n\" style=\"line-height:32px;\">\n<p class=\"tit\" style=\"line-height:25px;margin:10px\">Solu\u00e7\u00e3o<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\n<p><strong>(a)<\/strong> Seja a fun\u00e7\u00e3o [tex]f: A \\rightarrow   B[\/tex], com [tex]A,B \\subset \\mathbb{R} [\/tex], definida por [tex] f(x)=\\dfrac{x}{1-2^x}-\\dfrac{x}{2}.[\/tex]<br \/>\nDeterminar o dom\u00ednio m\u00e1ximo de [tex] f[\/tex] significa determinar o conjunto [tex]A [\/tex] formado por <strong>todos <\/strong>os n\u00fameros reais que podem ser colocados no lugar de [tex]x[\/tex] na lei de forma\u00e7\u00e3o [tex]y=f(x)[\/tex] de modo que, efetuados os c\u00e1lculos, obtenhamos um n\u00famero real [tex]y.[\/tex]<br \/>\nInicialmente, perceba que [tex]\\dfrac{x}{2}[\/tex] \u00e9 um n\u00famero real, independentemente do valor real que [tex]x[\/tex] assuma. Assim, para que [tex]\\dfrac{x}{1-2^x}-\\dfrac{x}{2}[\/tex] seja um n\u00famero real, basta que [tex]\\dfrac{x}{1-2^x}[\/tex] seja um n\u00famero real. Mas, sabemos que um quociente entre n\u00fameros reais n\u00e3o est\u00e1 definido somente se o seu denominador for igual a [tex]0;[\/tex] portanto, a \u00fanica restri\u00e7\u00e3o a ser imposta \u00e9 que [tex]1-2^x\\ne 0.[\/tex]<br \/>\nVamos ent\u00e3o determinar para quais valores de [tex]x[\/tex] temos [tex]1-2^x=0[\/tex] ou, de maneira equivalente,  [tex]2^x=1[\/tex]. Mas essa igualdade s\u00f3 ocorrer\u00e1 se [tex]x=0[\/tex]; portanto, o dom\u00ednio m\u00e1ximo da fun\u00e7\u00e3o [tex]f[\/tex] \u00e9 [tex] \\, \\fcolorbox{black}{#eee0e5}{$ A= \\mathbb{R}-\\{0\\}$} \\, [\/tex]. Em termos de intervalo, temos [tex] \\, \\fcolorbox{black}{#eee0e5}{$ A=\\left] -\\infty \\, , \\, 0\\right[ \\cup \\left] 0 \\, , \\, +\\infty \\, \\right[$} \\, .[\/tex]<\/p>\n<p style=\"margin:20px\">\n<p><strong>(b)<\/strong> Observe que, para  [tex]x \\in \\left] -\\infty \\, , \\, 0\\right[ \\cup \\left] 0 \\, , \\, +\\infty \\, \\right[[\/tex], segue que:<\/p>\n<table style=\"border:0px; text-align:justify;margin-left: auto; margin-right: auto;\">\n<tr>\n<td class=\"bf\">\n<p style=\"margin:10px\"> [tex]\\quad f(-x)=\\dfrac{-x}{1-2^{-x}}-\\dfrac{-x}{2}[\/tex]<\/p>\n<p style=\"margin:10px\"> [tex]\\quad f(-x)=\\dfrac{-x}{1-\\frac{1}{2^x}}+\\dfrac{x}{2}[\/tex]<\/p>\n<p style=\"margin:10px\"> [tex]\\quad f(-x)=\\dfrac{-x\\cdot 2^x}{2^x-1}+\\dfrac{x}{2}[\/tex]<\/p>\n<p style=\"margin:10px\"> [tex]\\quad f(-x)=\\dfrac{-2x\\cdot 2^x+ x \\cdot \\left(2^x-1\\right)}{2\\,\\left(2^x-1\\right)}[\/tex]<\/p>\n<p style=\"margin:10px\"> [tex]\\quad f(-x)=\\dfrac{-2x\\cdot 2^x+ x \\cdot 2^x-x}{2\\,\\left(2^x-1\\right)}[\/tex]<\/p>\n<p style=\"margin:10px\"> [tex]\\quad f(-x)=\\dfrac{-x\\cdot 2^x-x}{2\\,\\left(2^x-1\\right)}[\/tex]<\/p>\n<p style=\"margin:10px\"> [tex]\\quad f(-x)=\\dfrac{x\\cdot 2^x+x}{2\\,\\left(1-2^x\\right)}.\\qquad \\textcolor{#800000}{(i)}[\/tex]<\/p>\n<\/td>\n<td class=\"bf\" style=\"border-left: solid 2px #800000\">\n<p style=\"margin:15px\"> [tex]\\quad f(x)=\\dfrac{x}{1-2^{x}}-\\dfrac{x}{2}[\/tex]<\/p>\n<p style=\"margin:15px\"> [tex]\\quad f(x)=\\dfrac{2x- x \\cdot \\left(1-2^x\\right)}{2\\,\\left(1-2^x\\right)}[\/tex]<\/p>\n<p style=\"margin:15px\"> [tex]\\quad f(x)=\\dfrac{2x- x +x\\cdot 2^x}{2\\,\\left(1-2^x\\right)}[\/tex]<\/p>\n<p style=\"margin:15px\"> [tex]\\quad f(x)=\\dfrac{x +x\\cdot 2^x}{2\\,\\left(1-2^x\\right)}[\/tex]<\/p>\n<p style=\"margin:15px\"> [tex]\\quad f(x)=\\dfrac{x\\cdot 2^x+x}{2\\,\\left(1-2^x\\right)}.\\qquad \\textcolor{#800000}{(ii)}[\/tex]<\/p>\n<\/td>\n<\/tr>\n<\/table>\n<p>Por [tex]\\textcolor{#800000}{(i)}[\/tex] e [tex]\\textcolor{#800000}{(ii)}[\/tex], segue que [tex]f(x)=f(-x), \\forall x \\in \\left] -\\infty \\, , \\, 0\\right[ \\cup \\left] 0 \\, , \\, +\\infty \\, \\right[[\/tex], o que nos garante que [tex]f[\/tex] \u00e9 uma fun\u00e7\u00e3o par.<br \/>\nComo n\u00e3o temos [tex] f(-x)= -f(x) , \\forall x \\in \\left] -\\infty \\, , \\, 0\\right[ \\cup \\left] 0 \\, , \\, +\\infty \\, \\right[[\/tex], [tex]f[\/tex] n\u00e3o \u00e9 uma fun\u00e7\u00e3o \u00edmpar.<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 15px;border: 0 none\"\/>\n<p class=\"g2\" style=\"line-height:20px;\">Solu\u00e7\u00e3o elaborada pelos <strong>Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g3\">\nParticipou da discuss\u00e3o o Clube <strong>OCTETO MATEM\u00c1TICO<\/strong>.<\/div>\n<p style=\"margin:20px\">\n<table style=\" height:auto; margin-left: auto; margin-right: auto;width:640px;\">\n<tr>\n<td class=\"bf\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1001 alignright\" style=\"border: 0px;\" alt=\"\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2013\/01\/menino.jpg\" width=\"80\" height=\"115\" \/><\/td>\n<td class=\"bf\">\n<p class=\"fala\" style=\"width:520px;background-color: #FFF5EE; padding:10px; font-family: calibri; font-size: 13px;color: #1C1C1C; line-height: 20px; \"> Se for conveniente, voc\u00ea pode obter um arquivo PDF desta p\u00e1gina, com o problema e a solu\u00e7\u00e3o, clicando no bot\u00e3o abaixo.<br \/>\nVoc\u00ea pode abrir o arquivo diretamente no seu navegador (<em>Chrome<\/em>, <em>Edge<\/em>, <em>Firefox<\/em>, <em>Safari<\/em>, entre outros), mas tamb\u00e9m pode utilizar o software gratuito <em><strong>Adobe Acrobat Reader<\/strong><\/em>.<br \/>\nCaso o dispositivo que voc\u00ea est\u00e1 utilizando n\u00e3o tenha o <strong><em>Acrobat Reader<\/em><\/strong> instalado, \u00e9 s\u00f3 clicar <a href=\"http:\/\/get.adobe.com\/br\/reader\/\" rel=\"noopener\" target=\"_blank\"><strong>AQUI<\/strong><\/a> para fazer o download adequado ao seu dispositivo.<\/p>\n<\/td>\n<\/tr>\n<\/table>\n<div style=\" float:right;  width: 100px;\">\n<p style=\"text-align: center; border-style: dotted; border-color :#FFFAFA; border-width :thin; width:90px;  height:40px;  padding: 5px; background-color: #800000;\">\n<a href=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/12\/Fun\u00e7\u00e3o-par-fun\u00e7\u00e3o-\u00edmpar.pdf\" title=\" Arquivo PDF\" download=\"Clubes_ Fun\u00e7\u00e3o par, fun\u00e7\u00e3o \u00edmpar.pdf\"><span style=\"color: #FFFAFA; font-family: georgia; font-size: 16px;\"><strong>Arquivo <\/strong><\/span><\/a>\n <\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>&#9999; Link do problema para dispositivos da Apple. Problema (A partir da 2\u00aa s\u00e9rie do E. M. &#8211; N\u00edvel de dificuldade: M\u00e9dio) Uma fun\u00e7\u00e3o [tex]f: A \\rightarrow B[\/tex], com [tex]A,B \\subset \\mathbb{R} [\/tex], \u00e9 dita uma fun\u00e7\u00e3o par, se[tex] f(-x)=f(x)[\/tex], para todo [tex]x \\in A.[\/tex] Se [tex] f(-x)=-f(x)[\/tex], para todo [tex] x \\in A[\/tex], ent\u00e3o &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problema-para-ajudar-na-escola-funcao-par-funcao-impar\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-118819","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/118819","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=118819"}],"version-history":[{"count":62,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/118819\/revisions"}],"predecessor-version":[{"id":120398,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/118819\/revisions\/120398"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=118819"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}