{"id":205367,"date":"2023-11-10T05:22:34","date_gmt":"2023-11-10T08:22:34","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=205367"},"modified":"2023-11-10T05:22:34","modified_gmt":"2023-11-10T08:22:34","slug":"a-sala-3-problemas-e-videos","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/a-sala-3-problemas-e-videos\/","title":{"rendered":"(A) Sala 3 &#8211; Problemas e v\u00eddeos"},"content":{"rendered":"<p class=\"ti3\"  style=\"text-align:center;\">Clique no bot\u00e3o abaixo para visualizar a Sala.<\/p>\n<div style=\"text-align:center;margin-top:10px;\">\n<input class=\"bo1 c1 cb13\" style=\"width:150px;\" value=\"Sala\" title=\"Clique aqui\" type=\"button\" onclick=    \"if(this.parentNode.getElementsByTagName('div')[0].style.display != ''){this.parentNode.getElementsByTagName('div')[0].style.display = '';this.value = 'Ocultar';} else{this.parentNode.getElementsByTagName('div')[0].style.display = 'none'; this.value = 'Sala';}\" \/>  <\/p>\n<div style=\"display: none;\">\n<div class=\"gr1 cbm1\" style=\"color: #1e1e1e; font-size: 15px;\">\n<div class=\"nome cm\" style=\"font-size: 40px; line-height: 60px\"><strong>Problemas e v\u00eddeos<\/strong><\/div>\n<hr style=\"height:3px; width:100%;background-color:#8B5A2B;   margin-top: 8px; margin-bottom: 8px; border: 0 none\"\/>\n<hr style=\"height:3px; width:100%;background-color:#8B5A2B;   margin-top: 8px; margin-bottom: 20px; border: 0 none\"\/>\n<p style=\"margin:30px;font-size: 18px;\">\nNesta Sala voc\u00eas encontrar\u00e3o alguns problemas resolvidos para treinarmos o que foi estudado nas Salas anteriores.<br \/>\nMas que tal assistir primeiro alguns v\u00eddeos com aplica\u00e7\u00f5es de fun\u00e7\u00f5es e repassar um pouco da teoria??<\/p>\n<p class=\"ti2 cm\"  style=\"text-align:center; line-height: 28px; font-size: 25px;\">BOA DIVERS\u00c3O!!!!<\/p>\n<div class=\"gr1\" style=\"margin:30px;line-height:30px; width:85%; margin-left:auto; margin-right:auto; text-align:center; border: solid 2px #8B5A2B;font-size:16px \">\n<p class=\"ti1 cm \" style=\"line-height:30px;margin:10px\"><em>Carro Flex<\/em><\/p>\n<hr style=\"height:1px; width:100%;background-color:#8B5A2B; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nPara assistir, \u00e9 s\u00f3 clicar na setinha.<br \/>\n<iframe loading=\"lazy\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/UhIbDZaObfQ?rel=0\" allowfullscreen><\/iframe><br \/>\nV\u00eddeo da cole\u00e7\u00e3o de recursos educacionais da <a href=\"https:\/\/www.youtube.com\/@matematicamultimidia\" rel=\"noopener\" target=\"_blank\"><span style=\"color: #C68445;\"><strong>M\u00b3 Matem\u00e1tica Multim\u00eddia<\/strong><\/span><\/a>,<br \/>\ndesenvolvida pela Unicamp.<\/div>\n<p style=\"margin:30px;\">\n<div class=\"gr1\" style=\"margin:30px;line-height:30px; width:85%; margin-left:auto; margin-right:auto; text-align:center; border: solid 2px #8B5A2B;font-size:16px \">\n<p class=\"ti1 cm \" style=\"line-height:30px;margin:10px\"><em>Direitos do consumidor<\/em><\/p>\n<hr style=\"height:1px; width:100%;background-color:#8B5A2B; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nPara assistir, \u00e9 s\u00f3 clicar na setinha.<br \/>\n<iframe loading=\"lazy\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/ELdip5YBzso?rel=0\" allowfullscreen><\/iframe><br \/>\nV\u00eddeo da cole\u00e7\u00e3o de recursos educacionais da <a href=\"https:\/\/www.youtube.com\/@matematicamultimidia\" rel=\"noopener\" target=\"_blank\"><span style=\"color: #C68445;\"><strong>M\u00b3 Matem\u00e1tica Multim\u00eddia<\/strong><\/span><\/a>,<br \/>\ndesenvolvida pela Unicamp.<\/div>\n<p style=\"margin:30px;\">\n<div class=\"gr1\" style=\"margin:30px;line-height:30px; width:85%; margin-left:auto; margin-right:auto; text-align:center; border: solid 2px #8B5A2B;font-size:16px \">\n<p class=\"ti1 cm \" style=\"line-height:30px;margin:10px\"><em>Panquecas da dona Gl\u00f3ria<\/em><\/p>\n<hr style=\"height:1px; width:100%;background-color:#8B5A2B; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nPara assistir, \u00e9 s\u00f3 clicar na setinha.<br \/>\n<iframe loading=\"lazy\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/9yK-PJ_1wIw?rel=0\" allowfullscreen><\/iframe><br \/>\nV\u00eddeo da cole\u00e7\u00e3o de recursos educacionais da <a href=\"https:\/\/www.youtube.com\/@matematicamultimidia\" rel=\"noopener\" target=\"_blank\"><span style=\"color: #C68445;\"><strong>M\u00b3 Matem\u00e1tica Multim\u00eddia<\/strong><\/span><\/a>,<br \/>\ndesenvolvida pela Unicamp.<\/div>\n<p style=\"margin:30px;\">\n<div class=\"gr1\" style=\"margin:30px;line-height:30px; width:85%; margin-left:auto; margin-right:auto; text-align:center; border: solid 2px #8B5A2B;font-size:16px \">\n<p class=\"ti1 cm \" style=\"line-height:30px;margin:10px\"><em>O sonho<\/em><\/p>\n<hr style=\"height:1px; width:100%;background-color:#8B5A2B; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nPara assistir, \u00e9 s\u00f3 clicar na setinha.<br \/>\n<iframe loading=\"lazy\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/bHaGjxLdGzo?rel=0\" allowfullscreen><\/iframe><br \/>\nV\u00eddeo da cole\u00e7\u00e3o de recursos educacionais da <a href=\"https:\/\/www.youtube.com\/@matematicamultimidia\" rel=\"noopener\" target=\"_blank\"><span style=\"color: #C68445;\"><strong>M\u00b3 Matem\u00e1tica Multim\u00eddia<\/strong><\/span><\/a>,<br \/>\ndesenvolvida pela Unicamp.<\/div>\n<p style=\"margin:30px;\">\n<p class=\"ti2 cm\"  style=\"text-align:center; line-height: 28px; font-size: 25px;\">Um pouco mais de estudo!!!!<\/p>\n<div class=\"gr1\" style=\"margin:30px;line-height:30px; width:85%; margin-left:auto; margin-right:auto; text-align:center; border: solid 2px #8B5A2B;font-size:16px \">\n<p class=\"ti1 cm \" style=\"line-height:30px;margin:10px\"><em>N\u00fameros e Fun\u00e7\u00f5es Reais<\/em><\/p>\n<hr style=\"height:1px; width:100%;background-color:#8B5A2B; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nPara assistir, \u00e9 s\u00f3 clicar na setinha.<br \/>\n<iframe loading=\"lazy\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/_ysKNscRq0c?rel=0\" allowfullscreen><\/iframe><br \/>\nV\u00eddeo do PROFMAT, com o professor Elon Lages Lima.<\/div>\n<p style=\"margin:30px;\">\n<hr style=\"height:1px; width:100%;background-color:#8B5A2B;   margin-top: 30px; margin-bottom: 8px; border: 0 none\"\/>\n<hr style=\"height:6px; width:100%;background-color:#8B5A2B;  margin-top: 8px; margin-bottom: 8px;  border: 0 none\"\/>\n<hr style=\"height:1px; width:100%;background-color:#8B5A2B;   margin-top: 8px; margin-bottom: 30px; border: 0 none\"\/>\n<p style=\"margin:30px;font-size: 18px;\">\nAgora, vamos aos problemas!<\/p>\n<p class=\"ti2 cm\"  style=\"text-align:center; line-height: 28px; font-size: 22px;\">Mas, aten\u00e7\u00e3o&#8230;<br \/>\nN\u00e3o vale olhar a solu\u00e7\u00e3o antes de tentar resolver!\n<\/p>\n<p style=\"margin:30px;\">\n<div class=\"gr1 cbm5\" style=\"color: #1e1e1e; padding:20px; width:95%;font-size: 15px;\">\n<strong>Problema 1:<\/strong> (Adaptado &#8211; CEFET-MG  2013) Um experimento da \u00e1rea de Agronomia mostra que a temperatura m\u00ednima da superf\u00edcie do solo [tex]t(x),[\/tex] em [tex]^\\circ C,[\/tex] \u00e9 determinada em fun\u00e7\u00e3o do res\u00edduo [tex]x[\/tex] de planta e biomassa na superf\u00edcie, em [tex]g\/m^2[\/tex], conforme registrado na tabela abaixo.<\/p>\n<p style=\"margin:10px;\">[tex] \\qquad\\begin{array}{|c|c|c|c|c|c|c|c|}<br \/>\n\\hline<br \/>\nx\\;(g\/m^2)&#038;10&#038;20&#038;30&#038;40&#038;50&#038;60&#038;70\\\\<br \/>\n\\hline<br \/>\nt(x)\\;(^\\circ C) &#038; 7,24&#038; 7,30&#038; 7,36&#038; 7,42&#038; 7,48&#038; 7,54&#038; 7,60 \\\\<br \/>\n\\hline<br \/>\n\\end{array}[\/tex]<\/p>\n<p>Qual a lei de forma\u00e7\u00e3o da fun\u00e7\u00e3o [tex]t[\/tex]?<\/p>\n<p style=\"margin:20px;\">\n<div style=\"text-align:center;\"><input class=\"bo1 cm cbve3\" style=\"width:150px;\" value=\"Solu\u00e7\u00e3o\" title=\"Clique aqui\" type=\"button\" onclick=    \"if(this.parentNode.getElementsByTagName('div')[0].style.display != ''){this.parentNode.getElementsByTagName('div')[0].style.display = '';this.value = 'Ocultar';} else{this.parentNode.getElementsByTagName('div')[0].style.display = 'none'; this.value = 'Solu\u00e7\u00e3o';}\" \/>    <\/p>\n<div class=\"gr1 cbm3\" style=\"display: none; font-size: 15px; line-height:28px;\">\nObserve que os valores apresentados de [tex]x[\/tex] referentes a [tex]g\/m^2[\/tex] aumentam de [tex]10[\/tex] em [tex]10[\/tex], \u00e0 medida que os valores de [tex]t(x)[\/tex] aumentam de [tex]0,06[\/tex] em [tex]0,06[\/tex]. Portanto, os aumentos acontecem de forma linear, o que implica que [tex]t[\/tex] \u00e9 uma fun\u00e7\u00e3o afim.<br \/>\nConsidere [tex]t(x) = ax+b[\/tex], com [tex]a[\/tex] e [tex]b[\/tex] n\u00fameros reais.<br \/>\nEnt\u00e3o, substituindo as duas primeiras colunas de dados da tabela na lei de forma\u00e7\u00e3o de [tex]t(x)[\/tex], segue que<br \/>\n[tex]\\quad \\begin{cases}t(10) = 7,24\\\\<br \/>\nt(20) = 7,30,<br \/>\n\\end{cases}[\/tex]<br \/>\ndonde<br \/>\n[tex]\\quad \\begin{cases} 10a+b= 7,24\\\\<br \/>\n20a+b = 7,30.<br \/>\n\\end{cases}\\\\<br \/>\n~~[\/tex]<br \/>\nSubtraindo a primeira equa\u00e7\u00e3o da segunda, temos que:<br \/>\n[tex]\\quad 10a = 0,06\\\\<br \/>\n\\qquad  a = 0,006.[\/tex]<br \/>\nSubstituindo o valor de [tex]a[\/tex] na equa\u00e7\u00e3o [tex]10a+b= 7,24[\/tex], obtemos<br \/>\n[tex]\\quad b = 7,24-10a\\\\<br \/>\n\\quad b = 7,18.[\/tex]<br \/>\nLogo, [tex]~\\boxed{t(x) = 0,006x+7,18}.[\/tex]\n<\/div>\n<\/div>\n<\/div>\n<p style=\"margin:30px;\">\n<div class=\"gr1 cbm5\" style=\"color: #1e1e1e; padding:20px; width:95%;font-size: 15px;\">\n<strong>Problema 2:<\/strong> Sejam [tex]f[\/tex] e [tex]g[\/tex] fun\u00e7\u00f5es de [tex]\\mathbb{R}[\/tex] em [tex]\\mathbb{R},[\/tex] tais que [tex]Im(g)\\subset D(f).[\/tex]<br \/>\n Mostre que se [tex]f[\/tex] e [tex]g[\/tex] s\u00e3o injetivas, ent\u00e3o a fun\u00e7\u00e3o [tex]f \\circ g[\/tex] \u00e9 injetiva.<\/p>\n<p style=\"margin:20px;\">\n<div style=\"text-align:center;\"><input class=\"bo1 cm cbve3\" style=\"width:150px;\" value=\"Solu\u00e7\u00e3o\" title=\"Clique aqui\" type=\"button\" onclick=    \"if(this.parentNode.getElementsByTagName('div')[0].style.display != ''){this.parentNode.getElementsByTagName('div')[0].style.display = '';this.value = 'Ocultar';} else{this.parentNode.getElementsByTagName('div')[0].style.display = 'none'; this.value = 'Solu\u00e7\u00e3o';}\" \/>    <\/p>\n<div class=\"gr1 cbm3\" style=\"display: none; font-size: 15px; line-height:28px;\">\nSejam [tex]x[\/tex] e [tex]y[\/tex] n\u00fameros reais tais que [tex]\\boxed{(f \\circ g)(x) = (f \\circ g)(y)}.[\/tex]<br \/>\nEnt\u00e3o segue que:<br \/>\n[tex]\\quad(f \\circ g)(x) = (f \\circ g)(y)\\\\<br \/>\n\\quad f(g(x)) = f(g(y)).[\/tex]<br \/>\nComo [tex]f[\/tex] \u00e9 injetiva, ent\u00e3o [tex]g(x) = g(y).[\/tex]<br \/>\nAgora, usando o fato de que [tex]g[\/tex] \u00e9 injetiva, temos [tex]\\boxed{x = y}.[\/tex]<br \/>\n Portanto, [tex]f\\circ g[\/tex] \u00e9 injetiva.\n<\/div>\n<\/div>\n<\/div>\n<p style=\"margin:30px;\">\n<div class=\"gr1 cbm5\" style=\"color: #1e1e1e; padding:20px; width:95%;font-size: 15px;\">\n<strong>Problema 3:<\/strong> Sejam [tex]f[\/tex] e [tex]g[\/tex] fun\u00e7\u00f5es de [tex]\\mathbb{R}[\/tex] em [tex]\\mathbb{R},[\/tex] tais que [tex]Im(g)\\subset D(f).[\/tex]<br \/>\nMostre que se [tex]f[\/tex] e [tex]g[\/tex] s\u00e3o sobrejetivas, ent\u00e3o a fun\u00e7\u00e3o [tex]f \\circ g[\/tex] \u00e9 sobrejetiva.<\/p>\n<p style=\"margin:20px;\">\n<div style=\"text-align:center;\"><input class=\"bo1 cm cbve3\" style=\"width:150px;\" value=\"Solu\u00e7\u00e3o\" title=\"Clique aqui\" type=\"button\" onclick=    \"if(this.parentNode.getElementsByTagName('div')[0].style.display != ''){this.parentNode.getElementsByTagName('div')[0].style.display = '';this.value = 'Ocultar';} else{this.parentNode.getElementsByTagName('div')[0].style.display = 'none'; this.value = 'Solu\u00e7\u00e3o';}\" \/>    <\/p>\n<div class=\"gr1 cbm3\" style=\"display: none; font-size: 15px; line-height:28px;\">\nConsidere [tex]z\\in\\mathbb{R}.[\/tex] <\/p>\n<ul>\n<li>Como [tex]f[\/tex] \u00e9 sobrejetiva, existe [tex]y\\in\\mathbb{R}[\/tex] tal que [tex]f(y) = z.[\/tex] <\/li>\n<li>Agora, usando o fato de que [tex]g[\/tex] \u00e9 sobrejetiva, existe [tex]x\\in\\mathbb{R},[\/tex] tal que [tex]g(x) = y.[\/tex] <\/li>\n<\/ul>\n<p>Assim, mostramos que dado [tex]z\\in\\mathbb{R},[\/tex] existe [tex]x\\in \\mathbb{R}[\/tex] tal que [tex]f(g(x)) = z.[\/tex]<br \/>\nPortanto, [tex]f\\circ g[\/tex] \u00e9 sobrejetiva.\n<\/p><\/div>\n<\/div>\n<\/div>\n<p style=\"margin:30px;\">\n<div class=\"gr1 cbm5\" style=\"color: #1e1e1e; padding:20px; width:95%;font-size: 15px;\">\n<strong>Problema 4:<\/strong> A fun\u00e7\u00e3o [tex]f:\\mathbb{R}\\rightarrow\\mathbb{R}[\/tex] \u00e9 tal que [tex]f(1) = 2[\/tex], [tex]f(\\sqrt{2}) = 4[\/tex] e [tex]f(x+y) = f(x)f(y)[\/tex], para todos [tex]x, y\\in \\mathbb{R}[\/tex].<br \/>\nCalcule o valor de [tex]f(3+\\sqrt{2})[\/tex].<\/p>\n<p style=\"margin:20px;\">\n<div style=\"text-align:center;\"><input class=\"bo1 cm cbve3\" style=\"width:150px;\" value=\"Solu\u00e7\u00e3o\" title=\"Clique aqui\" type=\"button\" onclick=    \"if(this.parentNode.getElementsByTagName('div')[0].style.display != ''){this.parentNode.getElementsByTagName('div')[0].style.display = '';this.value = 'Ocultar';} else{this.parentNode.getElementsByTagName('div')[0].style.display = 'none'; this.value = 'Solu\u00e7\u00e3o';}\" \/>    <\/p>\n<div class=\"gr1 cbm3\" style=\"display: none; font-size: 15px; line-height:28px;\">\nPelas propriedades da fun\u00e7\u00e3o [tex]f[\/tex], temos o seguinte:<br \/>\n[tex]\\qquad\\begin{align}f(3+\\sqrt{2}) &#038;= f(3)\\cdot f(\\sqrt{2})\\\\<br \/>\n&#038;=f(3)\\cdot 4\\\\<br \/>\n&#038;=f(2+1)\\cdot 4\\\\<br \/>\n&#038;=f(2)\\cdot f(1)\\cdot 4\\\\<br \/>\n&#038;=f(2)\\cdot 2\\cdot 4\\\\<br \/>\n&#038;=f(1+1)\\cdot 8\\\\<br \/>\n&#038;=f(1)\\cdot f(1)\\cdot 8\\\\<br \/>\n&#038;=2\\cdot 2\\cdot 8\\\\<br \/>\n&#038;=32.<br \/>\n\\end{align}[\/tex]<\/div>\n<\/div>\n<\/div>\n<p style=\"margin:30px;\">\n<div class=\"gr1 cbm5\" style=\"color: #1e1e1e; padding:20px; width:95%;font-size: 15px;\">\n<strong>Problema 5:<\/strong> Dentre todos os ret\u00e2ngulos de mesmo per\u00edmetro, prove que o de maior \u00e1rea \u00e9 um quadrado.<\/p>\n<p style=\"margin:20px;\">\n<div style=\"text-align:center;\"><input class=\"bo1 cm cbve3\" style=\"width:150px;\" value=\"Solu\u00e7\u00e3o\" title=\"Clique aqui\" type=\"button\" onclick=    \"if(this.parentNode.getElementsByTagName('div')[0].style.display != ''){this.parentNode.getElementsByTagName('div')[0].style.display = '';this.value = 'Ocultar';} else{this.parentNode.getElementsByTagName('div')[0].style.display = 'none'; this.value = 'Solu\u00e7\u00e3o';}\" \/>    <\/p>\n<div class=\"gr1 cbm3\" style=\"display: none; font-size: 15px; line-height:28px;\">\nConsideremos todos os ret\u00e2ngulos de per\u00edmetro [tex]2p[\/tex].<br \/>\nSe, em um ret\u00e2ngulo gen\u00e9rico, um dos lados mede [tex]x[\/tex], ent\u00e3o a medida dos lados perpendiculares a esse \u00e9 [tex]p-x[\/tex]. Assim, a \u00e1rea desse ret\u00e2ngulo \u00e9 [tex]\\boxed{A(x) = x(p-x) = -x^2+px}.[\/tex]<br \/>\nVeja que [tex]A[\/tex] \u00e9 uma fun\u00e7\u00e3o quadr\u00e1tica com concavidade voltada para baixo, ent\u00e3o essa fun\u00e7\u00e3o admite valor m\u00e1ximo em<br \/>\n[tex]\\qquad x = -\\dfrac{b}{2a} = -\\dfrac{(p)}{2\\cdot (-1)} = \\dfrac{p}{2}[\/tex].<br \/>\nMas, sendo [tex]x=\\dfrac{p}{2}[\/tex], o ret\u00e2ngulo possui todos os lados de mesma medida, sendo portanto um quadrado.\n<\/div>\n<\/div>\n<\/div>\n<p style=\"margin:30px;\">\n<div class=\"gr1 cbm5\" style=\"color: #1e1e1e; padding:20px; width:95%;font-size: 15px;\">\n<strong>Problema 6:<\/strong> Seja [tex]f:\\mathbb{R}\\rightarrow\\mathbb{R}[\/tex] uma fun\u00e7\u00e3o \u00edmpar. Decida se a fun\u00e7\u00e3o [tex]f\\circ f[\/tex] \u00e9 par, \u00edmpar, ou nem par nem \u00edmpar.<\/p>\n<p style=\"margin:20px;\">\n<div style=\"text-align:center;\"><input class=\"bo1 cm cbve3\" style=\"width:150px;\" value=\"Solu\u00e7\u00e3o\" title=\"Clique aqui\" type=\"button\" onclick=    \"if(this.parentNode.getElementsByTagName('div')[0].style.display != ''){this.parentNode.getElementsByTagName('div')[0].style.display = '';this.value = 'Ocultar';} else{this.parentNode.getElementsByTagName('div')[0].style.display = 'none'; this.value = 'Solu\u00e7\u00e3o';}\" \/>    <\/p>\n<div class=\"gr1 cbm3\" style=\"display: none; font-size: 15px; line-height:28px;\">\nSe [tex]f[\/tex] \u00e9 uma fun\u00e7\u00e3o \u00edmpar, ent\u00e3o [tex]f(-x) = -f(x), \\forall~x \\in \\mathbb{R}.[\/tex]<br \/>\nAssim, se [tex]x \\in \\mathbb{R}[\/tex], segue que:<br \/>\n[tex]\\qquad (f\\circ f)(-x) = f(f(-x)) = f(-f(x)) = -f(f(x)) = -(f\\circ f)(x)[\/tex].<br \/>\nLogo, [tex]f\\circ f[\/tex] tamb\u00e9m \u00e9 uma fun\u00e7\u00e3o \u00edmpar.\n<\/div>\n<\/div>\n<\/div>\n<p style=\"margin:30px;\">\n<div class=\"gr1 cbm5\" style=\"color: #1e1e1e; padding:20px; width:95%;font-size: 15px;\">\n<strong>Problema 7:<\/strong> Seja [tex]f:\\mathbb{R}-\\{2\\}\\rightarrow\\mathbb{R}-\\{3\\}[\/tex] a fun\u00e7\u00e3o definida por [tex]f(x) =\\dfrac{3x-5}{x-2}[\/tex].<br \/>\nMostre que [tex]f[\/tex] \u00e9 uma bije\u00e7\u00e3o e obtenha a express\u00e3o para sua inversa.<\/p>\n<p style=\"margin:20px;\">\n<div style=\"text-align:center;\"><input class=\"bo1 cm cbve3\" style=\"width:150px;\" value=\"Solu\u00e7\u00e3o\" title=\"Clique aqui\" type=\"button\" onclick=    \"if(this.parentNode.getElementsByTagName('div')[0].style.display != ''){this.parentNode.getElementsByTagName('div')[0].style.display = '';this.value = 'Ocultar';} else{this.parentNode.getElementsByTagName('div')[0].style.display = 'none'; this.value = 'Solu\u00e7\u00e3o';}\" \/>    <\/p>\n<div class=\"gr1 cbm3\" style=\"display: none; font-size: 15px; line-height:33px;\">\nSejam [tex]x_1,x_2\\in D(f)[\/tex].<br \/>\nSe [tex]f(x_1) = f(x_2)[\/tex], ent\u00e3o temos que:<br \/>\n[tex]\\qquad \\dfrac{3x_1-5}{x_1-2} = \\dfrac{3x_2-5}{x_2-2}\\\\<br \/>\n\\qquad (x_2-2)(3x_1-5) = (x_1-2)(3x_2-5)\\\\<br \/>\n\\qquad 3x_1x_2-6x_1-5x_2+10 = 3x_1x_2-5x_1-6x_2+10\\\\<br \/>\n\\qquad  x_1=x_2.[\/tex]<br \/>\nPortanto, a fun\u00e7\u00e3o \u00e9 injetiva.<br \/>\nAl\u00e9m disso, a fun\u00e7\u00e3o [tex]f[\/tex] \u00e9 sobrejetora, ou seja, dado [tex]y\\in \\mathbb{R}-\\{3\\}[\/tex], existe [tex]x\\in\\mathbb{R}-\\{2\\}[\/tex] tal que [tex]f(x) = y[\/tex].<br \/>\nDe fato, considerando [tex]x = \\dfrac{5-2y}{3-y}[\/tex], temos<br \/>\n[tex]\\qquad \\begin{align}f(x) &#038;= f\\left(\\dfrac{5-2y}{3-y}\\right)\\\\<br \/>\n&#038;=\\dfrac{3\\left(\\frac{5-2y}{3-y}\\right)-5}{\\left(\\frac{5-2y}{3-y}\\right)-2}\\\\<br \/>\n&#038;=\\dfrac{\\left(\\frac{15-6y-5(3-y)}{3-y}\\right)}{\\left(\\frac{5-2y-2(3-y)}{3-y}\\right)}\\\\<br \/>\n&#038;=\\dfrac{\\left(\\frac{-y}{3-y}\\right)}{\\left(\\frac{-1}{3-y}\\right)}\\\\<br \/>\n&#038;=\\dfrac{-y}{-1}\\\\<br \/>\n&#038;=y.<br \/>\n\\end{align}[\/tex]<br \/>\nLogo, a fun\u00e7\u00e3o [tex]f[\/tex] \u00e9 bijetiva.<br \/>\nAgora vamos encontrar a fun\u00e7\u00e3o inversa de [tex]f[\/tex]. Sendo [tex]f(x)= y[\/tex], temos<br \/>\n[tex]\\qquad \\begin{align}<br \/>\n\\dfrac{3x-5}{x-2} &#038;= y\\\\<br \/>\n3x-5 &#038;= y(x-2)\\\\<br \/>\n3x-5 &#038;= xy-2y\\\\<br \/>\n3x-xy &#038;= 5-2y\\\\<br \/>\nx(3-y) &#038;= 5-2y\\\\<br \/>\nx &#038;= \\dfrac{5-2y}{3-y}\\\\<br \/>\nf^{-1}(y) &#038;= \\dfrac{5-2y}{3-y}.<br \/>\n\\end{align}[\/tex]<br \/>\nOu seja, a fun\u00e7\u00e3o inversa de [tex]f[\/tex] \u00e9 [tex]f^{-1}:\\mathbb{R}-\\{3\\}\\rightarrow \\mathbb{R}-\\{2\\}[\/tex] dada por [tex]f^{-1}(x) = \\dfrac{5-2x}{3-x}[\/tex].\n<\/div>\n<\/div>\n<\/div>\n<p style=\"margin:30px;\">\n<div class=\"gr1 cbm5\" style=\"color: #1e1e1e; padding:20px; width:95%;font-size: 15px;\">\n<strong>Problema 8:<\/strong> (UEL 2017) No plano cartesiano abaixo, cada um dos pontos representa a massa [tex](m)[\/tex] de um medicamento existente no sangue de um animal no instante [tex]t[\/tex] em que foi feita cada medi\u00e7\u00e3o depois do instante inicial, [tex]t=0[\/tex], da aplica\u00e7\u00e3o.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/12\/e4.png\" style=\"border:none\"alt=\"\" width=\"193\" height=\"321\" class=\"aligncenter size-full wp-image-174667\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/12\/e4.png 193w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/12\/e4-180x300.png 180w\" sizes=\"auto, (max-width: 193px) 100vw, 193px\" \/><br \/>\nConsiderando todos os instantes entre as medi\u00e7\u00f5es apresentadas no plano cartesiano, responda aos itens a seguir.<br \/>\n<strong>a)<\/strong> Sabendo que a rela\u00e7\u00e3o que descreve a massa [tex](m)[\/tex] do medicamento, ap\u00f3s [tex]t[\/tex] horas da aplica\u00e7\u00e3o, \u00e9 dada por [tex]m(t)=\\frac{C}{D+t}[\/tex] em que [tex]C[\/tex] e [tex]D[\/tex] s\u00e3o constantes, determine [tex]C[\/tex] e [tex]D[\/tex] na rela\u00e7\u00e3o dada.<br \/>\nJustifique sua resposta apresentando os c\u00e1lculos realizados na resolu\u00e7\u00e3o deste item.<br \/>\n<strong>b)<\/strong> Ap\u00f3s quanto tempo da administra\u00e7\u00e3o, a massa desse medicamento ser\u00e1 inferior a [tex]60\\%[\/tex] da massa que foi medida depois de [tex]2[\/tex] horas da aplica\u00e7\u00e3o?<br \/>\nJustifique sua resposta apresentando os c\u00e1lculos realizados na resolu\u00e7\u00e3o deste item. <\/p>\n<p style=\"margin:20px;\">\n<div style=\"text-align:center;\"><input class=\"bo1 cm cbve3\" style=\"width:150px;\" value=\"Solu\u00e7\u00e3o\" title=\"Clique aqui\" type=\"button\" onclick=    \"if(this.parentNode.getElementsByTagName('div')[0].style.display != ''){this.parentNode.getElementsByTagName('div')[0].style.display = '';this.value = 'Ocultar';} else{this.parentNode.getElementsByTagName('div')[0].style.display = 'none'; this.value = 'Solu\u00e7\u00e3o';}\" \/>    <\/p>\n<div class=\"gr1 cbm3\" style=\"display: none; font-size: 15px; line-height:28px;\">\n<strong>a)<\/strong> Observe que [tex]m(1) = 60[\/tex] e [tex]m(2) = 40[\/tex]. Assim,<br \/>\n[tex]\\qquad \\begin{cases}<br \/>\n\\frac{C}{D+1} = 60\\\\<br \/>\n\\frac{C}{D+2} = 40<br \/>\n\\end{cases}.[\/tex]<br \/>\nIsolando [tex]C[\/tex] em cada uma das equa\u00e7\u00f5es, temos [tex]C = 60(D+1)[\/tex] e [tex]C = 40(D+2)[\/tex]. Igualando as equa\u00e7\u00f5es, temos<br \/>\n[tex]\\qquad \\begin{align}<br \/>\n60(D+1) &#038;= 40(D+2)\\\\<br \/>\n60D+60 &#038;= 40D+80\\\\<br \/>\n20D &#038;= 20\\\\<br \/>\nD&#038;=1.\\end{align}[\/tex]<br \/>\nSubstituindo o valor de [tex]D[\/tex] na primeira express\u00e3o de [tex]C[\/tex], temos [tex]C=60(1+1)=120[\/tex]. Logo, [tex]C = 120[\/tex] e [tex]D = 1[\/tex].<br \/>\n<strong>b)<\/strong> Atrav\u00e9s do gr\u00e1fico, a partir de duas horas da aplica\u00e7\u00e3o, a massa do medicamento ser\u00e1 de [tex]40~mg[\/tex]. Queremos saber depois de quantas horas da aplica\u00e7\u00e3o, a massa ser\u00e1 [tex]60\\%[\/tex] da massa que foi medida depois de duas horas. Assim, queremos saber em qual instante a massa corresponde a [tex]60\\%[\/tex] de [tex]40 ~mg[\/tex]. Veja que [tex]\\dfrac{60}{100}\\cdot 40 = 24[\/tex].<br \/>\nLogo, observando o gr\u00e1fico, essa massa \u00e9 alcan\u00e7ada [tex]4[\/tex] horas depois da aplica\u00e7\u00e3o do medicamento.\n<\/div>\n<\/div>\n<\/div>\n<p style=\"margin:30px;\">\n<div class=\"gr1 cbm5\" style=\"color: #1e1e1e; padding:20px; width:95%;font-size: 15px;\">\n<strong>Problema 9:<\/strong> (G1 &#8211; IFSUL 2017 &#8211; Adaptada) Em uma disciplina, o n\u00famero de alunos reprovados por ano \u00e9 descrito pela fun\u00e7\u00e3o [tex]g(t)[\/tex], em que [tex]t[\/tex] \u00e9 dado em anos. Considerando [tex]f(g(t))=\\sqrt{2t+1}[\/tex] e [tex]f(t)=\\sqrt{t-2}[\/tex], como \u00e9 definida a fun\u00e7\u00e3o [tex]g(t)[\/tex]? <\/p>\n<p style=\"margin:20px;\">\n<div style=\"text-align:center;\"><input class=\"bo1 cm cbve3\" style=\"width:150px;\" value=\"Solu\u00e7\u00e3o\" title=\"Clique aqui\" type=\"button\" onclick=    \"if(this.parentNode.getElementsByTagName('div')[0].style.display != ''){this.parentNode.getElementsByTagName('div')[0].style.display = '';this.value = 'Ocultar';} else{this.parentNode.getElementsByTagName('div')[0].style.display = 'none'; this.value = 'Solu\u00e7\u00e3o';}\" \/>    <\/p>\n<div class=\"gr1 cbm3\" style=\"display: none; font-size: 15px; line-height:28px;\">\nSendo [tex]f(t)=\\sqrt{t-2}[\/tex], temos [tex]f(g(t))=\\sqrt{g(t)-2}[\/tex]. Mas, por outro lado, [tex]f(g(t))=\\sqrt{2t+1}[\/tex], ent\u00e3o<br \/>\n[tex]\\qquad \\begin{align}<br \/>\n\\sqrt{g(t)-2} &#038;= \\sqrt{2t+1}\\\\<br \/>\ng(t)-2 &#038;= 2t+1\\\\<br \/>\ng(t) &#038;= 2t+3.<br \/>\n\\end{align}[\/tex]\n<\/div>\n<\/div>\n<\/div>\n<p style=\"margin:30px;\">\n<div class=\"gr1 cbm5\" style=\"color: #1e1e1e; padding:20px; width:95%;font-size: 15px;\">\n<strong>Problema 10:<\/strong> (IME 2016 &#8211; Adaptada) Sejam as fun\u00e7\u00f5es [tex]f_n[\/tex], para [tex]n \\in \\{0,1,2,3,&#8230;\\}[\/tex], tais que [tex]f_0(x)=\\frac{1}{1-x}[\/tex] e [tex]f_n(x)=f_0(f_{n-1}(x))[\/tex], para [tex]n\\geq 1[\/tex]. Calcule [tex]f_{2023}(2023).[\/tex] <\/p>\n<p style=\"margin:20px;\">\n<div style=\"text-align:center;\"><input class=\"bo1 cm cbve3\" style=\"width:150px;\" value=\"Solu\u00e7\u00e3o\" title=\"Clique aqui\" type=\"button\" onclick=    \"if(this.parentNode.getElementsByTagName('div')[0].style.display != ''){this.parentNode.getElementsByTagName('div')[0].style.display = '';this.value = 'Ocultar';} else{this.parentNode.getElementsByTagName('div')[0].style.display = 'none'; this.value = 'Solu\u00e7\u00e3o';}\" \/>    <\/p>\n<div class=\"gr1 cbm3\" style=\"display: none; font-size: 15px; line-height:28px;\">\nAs fun\u00e7\u00e3o [tex]f_n[\/tex] foram definidas recursivamente como [tex]f_n(x)=f_0(f_{n-1}(x))[\/tex]. Ent\u00e3o,<br \/>\n<span style=\"color:#98622F; padding-left:18px;\">&#9679;<\/span>  [tex]f_0(x) = \\dfrac{1}{1-x}[\/tex];<br \/>\n<span style=\"color:#98622F; padding-left:18px;\">&#9679;<\/span>  [tex]f_1(x) = f_0(f_0(x)) = f_0\\left(\\frac{1}{1-x}\\right) = \\dfrac{1}{1-\\frac{1}{1-x}} = \\dfrac{1}{\\frac{1-x-1}{1-x}} = \\dfrac{1-x}{-x} = \\dfrac{x-1}{x}[\/tex];<br \/>\n<span style=\"color:#98622F; padding-left:18px;\">&#9679;<\/span>  [tex]f_2(x) = f_0(f_1(x)) = f_0(\\frac{x-1}{x}) = \\dfrac{1}{1-\\frac{x-1}{x}} = \\dfrac{1}{\\frac{x-x+1}{x}} = x[\/tex];<br \/>\n<span style=\"color:#98622F; padding-left:18px;\">&#9679;<\/span>  [tex]f_3(x) = f_0(f_2(x)) = f_0(x) = \\dfrac{1}{1-x}[\/tex].<br \/>\nObserve que [tex]f_3(x) = f_0(x)[\/tex], ent\u00e3o teremos [tex]f_4(x) = f_1(x), f_5(x) = f_2(x)[\/tex] e, depois, novamente, [tex]f_6(x) = f_3(x) = f_0(x)[\/tex].<br \/>\nComo podemos observar, sendo [tex]n[\/tex] m\u00faltiplo de [tex]3[\/tex], [tex]f_n(x) = f_0(x)[\/tex]. Portanto, [tex]f_{2022}(x) = f_0(x)[\/tex] e ent\u00e3o [tex]f_{2023}(x) = f_1(x) = \\dfrac{x-1}{x}[\/tex]. Logo, [tex]f_{2023}(2023) = \\dfrac{2023-1}{2023} = \\dfrac{2022}{2023}[\/tex].\n<\/div>\n<\/div>\n<\/div>\n<p style=\"margin:30px;\">\n<div class=\"gr1 cbm5\" style=\"color: #1e1e1e; padding:20px; width:95%;font-size: 15px;\">\n<strong>Problema 11:<\/strong> (ACAFE 2016 &#8211; Adaptada) O gr\u00e1fico a seguir representa a fun\u00e7\u00e3o real [tex]f(x)[\/tex] definida no intervalo [tex][-1,6][\/tex].<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/12\/e18.png\" style=\"border:none\" alt=\"\" width=\"250\" height=\"166\" class=\"aligncenter size-full wp-image-174676\" \/><br \/>\nConsiderando a fun\u00e7\u00e3o [tex]h(x)=f(x-2)[\/tex], qual o valor da express\u00e3o dada por [tex]f(h(3))+f(f(4))[\/tex]?<\/p>\n<p style=\"margin:20px;\">\n<div style=\"text-align:center;\"><input class=\"bo1 cm cbve3\" style=\"width:150px;\" value=\"Solu\u00e7\u00e3o\" title=\"Clique aqui\" type=\"button\" onclick=    \"if(this.parentNode.getElementsByTagName('div')[0].style.display != ''){this.parentNode.getElementsByTagName('div')[0].style.display = '';this.value = 'Ocultar';} else{this.parentNode.getElementsByTagName('div')[0].style.display = 'none'; this.value = 'Solu\u00e7\u00e3o';}\" \/>    <\/p>\n<div class=\"gr1 cbm3\" style=\"display: none; font-size: 15px; line-height:28px;\">\n<p>Observando o gr\u00e1fico, temos<br \/>\n[tex]\\qquad\\begin{align}f(h(3))+f(f(4)) &#038;= f(f(3-2))+f(1)\\\\<br \/>\n&#038;= f(f(1))+4\\\\<br \/>\n&#038;= f(4)+4\\\\<br \/>\n&#038;= 1+4\\\\<br \/>\n&#038;= 5.<br \/>\n\\end{align}[\/tex]\n<\/p><\/div>\n<\/div>\n<\/div>\n<p style=\"margin:30px;\">\n<hr style=\"height:3px; width:100%;background-color:#8B5A2B;  margin-top: 40px; margin-bottom: 8px;  border: 0 none\"\/>\n<hr style=\"height:3px; width:100%;background-color:#8B5A2B;  margin-top: 8px; margin-bottom: 20px;  border: 0 none\"\/>\n<p style=\"text-align: right;\"><span style=\"color: #8B5A2B; font-size: 18px;\"><strong>Equipe COM &#8211; OBMEP<\/strong><\/span><\/p>\n<\/div>\n<p class=\"fim\"><a title=\"Voltar.\" href=\"http:\/\/clubes.obmep.org.br\/blog\/?p=205334\"><span style=\"color: #C68445;\"><strong>Voltar para Sala Inicial<\/strong><\/span><\/a><\/p>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Clique no bot\u00e3o abaixo para visualizar a Sala. Problemas e v\u00eddeos Nesta Sala voc\u00eas encontrar\u00e3o alguns problemas resolvidos para treinarmos o que foi estudado nas Salas anteriores. Mas que tal assistir primeiro alguns v\u00eddeos com aplica\u00e7\u00f5es de fun\u00e7\u00f5es e repassar um pouco da teoria?? BOA DIVERS\u00c3O!!!! Carro Flex Para assistir, \u00e9 s\u00f3 clicar na setinha. &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/a-sala-3-problemas-e-videos\/\">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-205367","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/205367","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=205367"}],"version-history":[{"count":2,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/205367\/revisions"}],"predecessor-version":[{"id":205369,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/205367\/revisions\/205369"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=205367"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}