{"id":11835,"date":"2013-12-03T01:38:55","date_gmt":"2013-12-03T03:38:55","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=11835"},"modified":"2023-01-08T05:59:38","modified_gmt":"2023-01-08T08:59:38","slug":"problemao-equacao-funcional","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problemao-equacao-funcional\/","title":{"rendered":".Problem\u00e3o: Equa\u00e7\u00e3o Funcional"},"content":{"rendered":"<div style=\"border: 3px solid #800000; padding: 8px;text-align: justify; padding-left: 10px; padding-right: 10px;font-family: verdana; color: #616161; line-height: 35px;\">\n<p class=\"tit\" style=\"line-height:25px;margin:10px\">Problema<br \/>\n<span style=\"font-size:12px;color:#800000\"> (Indicado a partir do 1\u00ba ano do E. M.)<\/span><\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\n<span style=\"font-size: 14px;\">(R\u00fassia)<\/span> Seja a fun\u00e7\u00e3o dada por [tex]f(x)=x^2+12x+30[\/tex].<br \/>\nResolver a equa\u00e7\u00e3o [tex]f(f(f(f(f(x)))))=0[\/tex].<\/div>\n<p>&nbsp;<\/p>\n<div style=\"border: 3px  solid  #800000; padding: 8px;\">\n<p class=\"tit\" style=\"line-height:25px;margin:10px\"><strong>Solu\u00e7\u00e3o<\/strong><\/span><\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\n<p style=\"text-align: justify; padding: 5px;padding-left: 12px;font-family: verdana; color: #616161; line-height: 32px;\">\nInicialmente, veja que<br \/>\n[tex]\\qquad f(x)=x^2+12x+30\\\\<br \/>\n\\qquad f(x)=x^2+12x+30+6-6\\\\<br \/>\n\\qquad f(x)=x^2+12x+36-6\\\\<br \/>\n\\qquad \\boxed{f(x)=(x+6)^2-6}.[\/tex]<br \/>\nDeste modo:<br \/>\n[tex]\\qquad f(\\textcolor{red}{f(x)})=(\\textcolor{red}{f(x)}+6)^2-6\\\\<br \/>\n\\qquad f(\\textcolor{red}{f(x)})=\\left(\\textcolor{red}{(x+6)^2-6}+6\\right)^2-6\\\\<br \/>\n\\qquad f(f(x))=\\left((x+6)^2\\right)^2-6\\\\<br \/>\n\\qquad \\boxed{f(f(x))=(x+6)^4-6}.[\/tex]<br \/>\nCom isso:<br \/>\n[tex]\\qquad f(f(\\textcolor{red}{f(x)}))=(\\textcolor{red}{f(x)}+6)^4-6\\\\<br \/>\n\\qquad f(f(\\textcolor{red}{f(x)}))=\\left(\\textcolor{red}{(x+6)^2-6}+6\\right)^4-6\\\\<br \/>\n\\qquad f(f(f(x)))=\\left((x+6)^2\\right)^4-6\\\\<br \/>\n\\qquad \\boxed{f(f(f(x)))=\\left(x+6\\right)^8-6}.[\/tex]<br \/>\nA partir do exposto, podemos conjecturar que<br \/>\n[tex]\\qquad f(f(f(f(f(x)))))=(x+6)^{32}-6[\/tex]. (Mostre isso!)<br \/>\nAssim, [tex]f(f(f(f(f(x)))))=(x+6)^{32}-6=0[\/tex] e, portanto, segue que:<br \/>\n[tex]\\qquad (x+6)^{32}-6=0[\/tex]<br \/>\n[tex]\\qquad (x+6)^{32}=6[\/tex]<br \/>\n[tex]\\qquad x+6=\\pm \\sqrt[32]{6}[\/tex]<br \/>\n[tex]\\qquad \\fcolorbox{black}{#eee0e5}{$x=-6\\pm \\sqrt[32]{6}$}\\,.[\/tex]<\/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","protected":false},"excerpt":{"rendered":"<p>Problema (Indicado a partir do 1\u00ba ano do E. M.) (R\u00fassia) Seja a fun\u00e7\u00e3o dada por [tex]f(x)=x^2+12x+30[\/tex]. Resolver a equa\u00e7\u00e3o [tex]f(f(f(f(f(x)))))=0[\/tex]. &nbsp; Solu\u00e7\u00e3o Inicialmente, veja que [tex]\\qquad f(x)=x^2+12x+30\\\\ \\qquad f(x)=x^2+12x+30+6-6\\\\ \\qquad f(x)=x^2+12x+36-6\\\\ \\qquad \\boxed{f(x)=(x+6)^2-6}.[\/tex] Deste modo: [tex]\\qquad f(\\textcolor{red}{f(x)})=(\\textcolor{red}{f(x)}+6)^2-6\\\\ \\qquad f(\\textcolor{red}{f(x)})=\\left(\\textcolor{red}{(x+6)^2-6}+6\\right)^2-6\\\\ \\qquad f(f(x))=\\left((x+6)^2\\right)^2-6\\\\ \\qquad \\boxed{f(f(x))=(x+6)^4-6}.[\/tex] Com isso: [tex]\\qquad f(f(\\textcolor{red}{f(x)}))=(\\textcolor{red}{f(x)}+6)^4-6\\\\ \\qquad f(f(\\textcolor{red}{f(x)}))=\\left(\\textcolor{red}{(x+6)^2-6}+6\\right)^4-6\\\\ \\qquad f(f(f(x)))=\\left((x+6)^2\\right)^4-6\\\\ \\qquad \\boxed{f(f(f(x)))=\\left(x+6\\right)^8-6}.[\/tex] A &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problemao-equacao-funcional\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":164,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-11835","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/11835","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\/164"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=11835"}],"version-history":[{"count":17,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/11835\/revisions"}],"predecessor-version":[{"id":11841,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/11835\/revisions\/11841"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=11835"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}