{"id":11805,"date":"2014-02-24T00:58:09","date_gmt":"2014-02-24T00:58:09","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=11805"},"modified":"2022-01-21T16:12:15","modified_gmt":"2022-01-21T16:12:15","slug":"tres-funcoes-f-g-e-fracfg","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=11805","title":{"rendered":"Tr\u00eas fun\u00e7\u00f5es: $f$, $g$ e $\\frac{f}{g}$"},"content":{"rendered":"<p><ul id='GTTabs_ul_11805' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_11805' class='GTTabs_curr'><a  id=\"11805_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_11805' ><a  id=\"11805_1\" onMouseOver=\"GTTabsShowLinks('Resolu\u00e7\u00e3o'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Resolu\u00e7\u00e3o<\/a><\/li>\n<\/ul>\n\n<div class='GTTabs_divs GTTabs_curr_div' id='GTTabs_0_11805'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><br \/>\nSejam $f$ e $g$ duas fun\u00e7\u00f5es definidas por: \\[\\begin{array}{*{20}{c}}<br \/>\n{f\\left( x \\right) = {x^2} &#8211; 4}&amp;{\\text{e}}&amp;{g\\left( x \\right) = x + 2}<br \/>\n\\end{array}\\]<\/p>\n<p>Caracterize a fun\u00e7\u00e3o $\\frac{f}{g}$ e estude o seu sinal, relacionando-o com o sinal quer da fun\u00e7\u00e3o $f$ quer da fun\u00e7\u00e3o $g$.<\/p>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_11805' onClick='GTTabs_show(1,11805)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_11805'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<blockquote>\n<p>Sejam $f$ e $g$ duas fun\u00e7\u00f5es definidas por: \\[\\begin{array}{*{20}{c}}<br \/>\n{f\\left( x \\right) = {x^2} &#8211; 4}&amp;{\\text{e}}&amp;{g\\left( x \\right) = x + 2}<br \/>\n\\end{array}\\]<\/p>\n<p>Caracterize a fun\u00e7\u00e3o $\\frac{f}{g}$ e estude o seu sinal, relacionando-o com o sinal quer da fun\u00e7\u00e3o $f$ quer da fun\u00e7\u00e3o $g$.<\/p>\n<\/blockquote>\n<p>\u00ad<\/p>\n<p>Comecemos por determinar o dom\u00ednio de $\\frac{f}{g}$:<\/p>\n<p>\\[\\begin{array}{*{20}{l}}<br \/>\n{{D_{\\frac{f}{g}}}}&amp; = &amp;{{D_f} \\cap {D_g} \\cap \\left\\{ {x \\in \\mathbb{R}:g\\left( x \\right) \\ne 0} \\right\\}} \\\\<br \/>\n{}&amp; = &amp;{\\mathbb{R} \\cap \\mathbb{R} \\cap \\left\\{ {x \\in \\mathbb{R}:x + 2 \\ne 0} \\right\\}} \\\\<br \/>\n{}&amp; = &amp;{\\mathbb{R}\\backslash \\left\\{ { &#8211; 2} \\right\\}}<br \/>\n\\end{array}\\]<\/p>\n<p>\u00a0Determinemos, agora, $\\frac{f}{g}\\left( x \\right)$:<\/p>\n<\/p>\n<p>\\[\\frac{f}{g}\\left( x \\right) = \\frac{{f\\left( x \\right)}}{{g\\left( x \\right)}} = \\frac{{{x^2} &#8211; 4}}{{x + 2}} = \\frac{{\\left( {x + 2} \\right)\\left( {x &#8211; 2} \\right)}}{{x + 2}} = x &#8211; 2,\\,\\forall x \\in \\mathbb{R}\\backslash \\left\\{ { &#8211; 2} \\right\\}\\]Logo:<\/p>\n<p>\\[\\begin{array}{*{20}{r}}<br \/>\n{\\frac{f}{g}:}&amp;{\\mathbb{R}\\backslash \\left\\{ { &#8211; 2} \\right\\} \\to \\mathbb{R}} \\\\<br \/>\n{}&amp;{x \\to x &#8211; 2}<br \/>\n\\end{array}\\]<\/p>\n<\/p>\n<table class=\" aligncenter\" style=\"width: 80%;\" border=\"0\" align=\"center\">\n<tbody>\n<tr>\n<td style=\"border: 1px solid #778899; width: 100px; text-align: center;\">$x$<\/td>\n<td style=\"border: 1px solid #778899; width: 100px; text-align: left;\">$ &#8211; \\infty $<\/td>\n<td style=\"border: 1px solid #778899; width: 50px; text-align: center;\">$-2$<\/td>\n<td style=\"border: 1px solid #778899; width: 100px;\"><\/td>\n<td style=\"border: 1px solid #778899; width: 50px; text-align: center;\">$2$<\/td>\n<td style=\"border: 1px solid #778899; width: 100px; text-align: right;\">$ + \\infty $<\/td>\n<\/tr>\n<tr>\n<td style=\"border: 1px solid #778899; width: 100px; text-align: center;\">$\\frac{f}{g}\\left( x \\right) = x &#8211; 2$<\/td>\n<td style=\"border: 1px solid #778899; width: 100px; text-align: center;\">$-$<\/td>\n<td style=\"border: 1px solid #778899; width: 50px; text-align: center;\">n.d.<\/td>\n<td style=\"border: 1px solid #778899; width: 100px; text-align: center;\">$-$<\/td>\n<td style=\"border: 1px solid #778899; width: 50px; text-align: center;\">$0$<\/td>\n<td style=\"border: 1px solid #778899; width: 100px; text-align: center;\">$+$<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ul>\n<li><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/02\/11-2-pag125-10.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"11806\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=11806\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/02\/11-2-pag125-10.png\" data-orig-size=\"1394,1286\" data-comments-opened=\"1\" data-image-meta=\"{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;0&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}\" data-image-title=\"Gr\u00e1ficos\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/02\/11-2-pag125-10-1024x944.png\" class=\"alignright  wp-image-11806\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/02\/11-2-pag125-10.png\" alt=\"Gr\u00e1ficos\" width=\"402\" height=\"370\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/02\/11-2-pag125-10.png 1394w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/02\/11-2-pag125-10-300x276.png 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/02\/11-2-pag125-10-1024x944.png 1024w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/02\/11-2-pag125-10-150x138.png 150w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/02\/11-2-pag125-10-400x369.png 400w\" sizes=\"auto, (max-width: 402px) 100vw, 402px\" \/><\/a>${\\frac{f}{g}}$ \u00e9 negativa no intervalo $\\left] { &#8211; \\infty , &#8211; 2} \\right[$, pois neste intervalo $f$ e $g$ s\u00e3o de sinais contr\u00e1rios;<\/li>\n<li>${\\frac{f}{g}}$ n\u00e3o est\u00e1 definida em $x = &#8211; 2$, pois este valor \u00e9 zero de $g$;<\/li>\n<li>${\\frac{f}{g}}$ \u00e9 negativa no intervalo $\\left] { &#8211; 2,2} \\right[$, pois neste intervalo $f$ e $g$ s\u00e3o de sinais contr\u00e1rios;<\/li>\n<li>${\\frac{f}{g}}$ \u00e9 nula em $x = 2$, pois este valor \u00e9 zero de $f$ e $g\\left( 2 \\right) &gt; 0$;<\/li>\n<li>${\\frac{f}{g}}$ \u00e9\u00a0positiva no intervalo $\\left] {2, + \\infty } \\right[$, pois neste intervalo $f$ e $g$ s\u00e3o do mesmo sinal.<\/li>\n<\/ul>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_11805' onClick='GTTabs_show(0,11805)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Sejam $f$ e $g$ duas fun\u00e7\u00f5es definidas por: \\[\\begin{array}{*{20}{c}} {f\\left( x \\right) = {x^2} &#8211; 4}&amp;{\\text{e}}&amp;{g\\left( x \\right) = x + 2} \\end{array}\\] Caracterize a fun\u00e7\u00e3o $\\frac{f}{g}$ e estude o seu&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":20821,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[98,97,147],"tags":[422,364],"series":[],"class_list":["post-11805","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-11--ano","category-aplicando","category-operacoes-com-funcoes","tag-11-o-ano","tag-funcao-quociente"],"views":1879,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/02\/11V2Pag125-10_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/11805","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=11805"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/11805\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/20821"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=11805"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=11805"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=11805"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=11805"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}