{"id":7960,"date":"2012-04-03T16:10:46","date_gmt":"2012-04-03T15:10:46","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=7960"},"modified":"2022-01-13T22:56:44","modified_gmt":"2022-01-13T22:56:44","slug":"estude-e-represente-graficamente-a-funcao","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=7960","title":{"rendered":"Estude e represente graficamente a fun\u00e7\u00e3o"},"content":{"rendered":"<p><ul id='GTTabs_ul_7960' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_7960' class='GTTabs_curr'><a  id=\"7960_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_7960' ><a  id=\"7960_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_7960'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Estude e represente graficamente a fun\u00e7\u00e3o seno hiperb\u00f3lico\u00a0definida em $\\mathbb{R}$ por: $$f(x) = {\\text{senh}}\\,x = \\frac{{{e^x} &#8211; {e^{ &#8211; x}}}}{2}$$<\/p>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_7960' onClick='GTTabs_show(1,7960)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_7960'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<blockquote>\n<p>$$f(x) = {\\text{senh}}\\,x = \\frac{{{e^x} &#8211; {e^{ &#8211; x}}}}{2}$$<\/p>\n<\/blockquote>\n<ul style=\"list-style-type: square;\">\n<li><strong>Dom\u00ednio<\/strong><\/li>\n<\/ul>\n<p>\u00a0${D_f} = \\mathbb{R}$<\/p>\n<ul style=\"list-style-type: square;\">\n<li><strong>Zeros<\/strong><\/li>\n<\/ul>\n<p>$$\\begin{array}{*{20}{l}}<br \/>\n{f(x) = 0}&amp; \\Leftrightarrow &amp;{\\frac{{{e^x} &#8211; {e^{ &#8211; x}}}}{2} = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{{e^x} &#8211; {e^{ &#8211; x}} = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{{e^x} = {e^{ &#8211; x}}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x =\u00a0 &#8211; x} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = 0}<br \/>\n\\end{array}$$<\/p>\n<p>A fun\u00e7\u00e3o tem um zero: $x=0$.<\/p>\n<ul style=\"list-style-type: square;\">\n<li><strong>Paridade<\/strong><\/li>\n<\/ul>\n<p>$$f( &#8211; x) = \\frac{{{e^{ &#8211; x}} &#8211; {e^x}}}{2} =\u00a0 &#8211; \\frac{{{e^x} &#8211; {e^{ &#8211; x}}}}{2} =\u00a0 &#8211; f(x),\\forall x \\in {D_f}$$<\/p>\n<p>A fun\u00e7\u00e3o \u00e9 <strong>\u00edmpar<\/strong>, pois $f( &#8211; x) =\u00a0 &#8211; f(x),\\forall x \\in {D_f}$.<\/p>\n<ul style=\"list-style-type: square;\">\n<li><strong>Ass\u00edntotas<\/strong><\/li>\n<\/ul>\n<p>A fun\u00e7\u00e3o \u00e9 cont\u00ednua em $\\mathbb{R}$, logo o seu gr\u00e1fico n\u00e3o tem ass\u00edntotas verticais.<\/p>\n<p>$$\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } f(x) = \\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\frac{{{e^x} &#8211; {e^{ &#8211; x}}}}{2} = \\frac{1}{2} \\times \\left( {\\underbrace {\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } {e^x}}_{ + \\infty } &#8211; \\underbrace {\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } {e^{ &#8211; x}}}_0} \\right) =\u00a0 + \\infty $$<\/p>\n<p>Como a fun\u00e7\u00e3o \u00e9 \u00edmpar, ent\u00e3o $\\mathop {\\lim }\\limits_{x \\to\u00a0 &#8211; \\infty } f(x) =\u00a0 &#8211; \\infty $.<\/p>\n<p>Logo, o gr\u00e1fico da fun\u00e7\u00e3o n\u00e3o tem ass\u00edntotas horizontais.<\/p>\n<p>$$m = \\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\frac{{f(x)}}{x} = \\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\frac{{{e^x} &#8211; {e^{ &#8211; x}}}}{{2x}} = \\frac{1}{2} \\times \\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\frac{{{e^x}\\left( {1 &#8211; {e^{ &#8211; 2x}}} \\right)}}{x} = \\frac{1}{2} \\times \\underbrace {\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\frac{{{e^x}}}{x}}_{ + \\infty } \\times \\underbrace {\\mathop {\\lim }\\limits_{x \\to\u00a0 + \\infty } \\left( {1 &#8211; {e^{ &#8211; 2x}}} \\right)}_1 =\u00a0 + \\infty $$<\/p>\n<p>Portanto, o gr\u00e1fico da fun\u00e7\u00e3o tamb\u00e9m n\u00e3o tem ass\u00edntotas obl\u00edquas.<\/p>\n<ul style=\"list-style-type: square;\">\n<li><strong>Monotonia e extremos<\/strong><\/li>\n<\/ul>\n<p>Ora, \\[f&#8217;\\left( x \\right) = {\\left( {\\frac{{{e^x} &#8211; {e^{ &#8211; x}}}}{2}} \\right)^\\prime } = \\frac{{{e^x} + {e^{ &#8211; x}}}}{2}\\]<\/p>\n<p>Como $f'(x) &gt; 0,\\forall x \\in {D_f}$, ent\u00e3o a fun\u00e7\u00e3o \u00e9 estritamente crescente no seu dom\u00ednio.<\/p>\n<p>\\[f&#8221;\\left( x \\right) = {\\left( {\\frac{{{e^x} + {e^{ &#8211; x}}}}{2}} \\right)^\\prime } = \\frac{{{e^x} &#8211; {e^{ &#8211; x}}}}{2} = \\frac{{{e^x}\\left( {1 &#8211; {e^{ &#8211; 2x}}} \\right)}}{2} = f(x)\\]<\/p>\n<\/p>\n<table class=\" aligncenter\" style=\"width: 80%;\" border=\"0\" align=\"center\">\n<tbody>\n<tr>\n<td style=\"text-align: center; border: #0000ff 1px solid;\">$x$<\/td>\n<td style=\"text-align: left; border: #0000ff 1px solid;\">${ &#8211; \\infty }$<\/td>\n<td style=\"text-align: center; border: #0000ff 1px solid;\">$0$<\/td>\n<td style=\"text-align: right; border: #0000ff 1px solid;\">\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 ${ + \\infty }$<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center; border: #0000ff 1px solid;\">$\\frac{{{e^x}}}{2}$<\/td>\n<td style=\"text-align: center; border: #0000ff 1px solid;\">$+$<\/td>\n<td style=\"text-align: center; border: #0000ff 1px solid;\">$+$<\/td>\n<td style=\"text-align: center; border: #0000ff 1px solid;\">$+$<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center; border: #0000ff 1px solid;\">${1 &#8211; {e^{ &#8211; 2x}}}$<\/td>\n<td style=\"text-align: center; border: #0000ff 1px solid;\">$-$<\/td>\n<td style=\"text-align: center; border: #0000ff 1px solid;\">$0$<\/td>\n<td style=\"text-align: center; border: #0000ff 1px solid;\">$+$<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center; border: #0000ff 1px solid;\">$f&#8221;(x) = \\frac{{{e^x}\\left( {1 &#8211; {e^{ &#8211; 2x}}} \\right)}}{2}$<\/td>\n<td style=\"text-align: center; border: #0000ff 1px solid;\">$-$<\/td>\n<td style=\"text-align: center; border: #0000ff 1px solid;\">$0$<\/td>\n<td style=\"text-align: center; border: #0000ff 1px solid;\">$+$<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center; border: #0000ff 1px solid;\">Concavidade do gr\u00e1fico de $f$<\/td>\n<td style=\"text-align: center; border: #0000ff 1px solid;\">$ \\cap $<\/td>\n<td style=\"text-align: center; border: #0000ff 1px solid;\">P.I.<\/td>\n<td style=\"text-align: center; border: #0000ff 1px solid;\">$ \\cup $<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ul style=\"list-style-type: square;\">\n<li><strong>Representa\u00e7\u00e3o 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Enunciado<\/p>\n","protected":false},"author":1,"featured_media":14083,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[226,97,292],"tags":[427,145],"series":[],"class_list":["post-7960","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-12--ano","category-aplicando","category-calculo-diferencial","tag-12-o-ano","tag-derivadas-2"],"views":2959,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/03\/Mat28.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/7960","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=7960"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/7960\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/14083"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=7960"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=7960"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=7960"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=7960"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}