{"id":11857,"date":"2014-04-30T15:53:36","date_gmt":"2014-04-30T14:53:36","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=11857"},"modified":"2022-01-23T15:04:12","modified_gmt":"2022-01-23T15:04:12","slug":"a-inversa-de-uma-funcao","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=11857","title":{"rendered":"A inversa de uma fun\u00e7\u00e3o"},"content":{"rendered":"<p><ul id='GTTabs_ul_11857' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_11857' class='GTTabs_curr'><a  id=\"11857_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_11857' ><a  id=\"11857_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_11857'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>A fun\u00e7\u00e3o $f$ tem dom\u00ednio $\\left[ {0, + \\infty } \\right[$ e \u00e9 definida por $f\\left( x \\right) = 4{x^2} + 1$.<\/p>\n<ol>\n<li>Esboce o gr\u00e1fico de $f$ e indique o contradom\u00ednio da fun\u00e7\u00e3o.<\/li>\n<li>Explique porque existe inversa de $f$ e determine uma express\u00e3o para ${f^{ &#8211; 1}}\\left( x \\right)$.<\/li>\n<li>Sabendo que $g$ \u00e9 outra fun\u00e7\u00e3o cujo dom\u00ednio \u00e9 $\\left[ {0, + \\infty } \\right[$ e \u00e9 definida por $g\\left( x \\right) = \\sqrt {x + 6} $, resolva a desigualdade $f\\left( {g\\left( x \\right)} \\right) \\geqslant f\\left( x \\right)$.<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_11857' onClick='GTTabs_show(1,11857)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_11857'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"11860\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=11860\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/04\/11-2-pag174-15b.png\" data-orig-size=\"397,470\" 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\/04\/11-2-pag174-15b.png\" class=\"alignright size-full wp-image-11860\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/04\/11-2-pag174-15b.png\" alt=\"Gr\u00e1ficos\" width=\"397\" height=\"470\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/04\/11-2-pag174-15b.png 397w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/04\/11-2-pag174-15b-253x300.png 253w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/04\/11-2-pag174-15b-126x150.png 126w\" sizes=\"auto, (max-width: 397px) 100vw, 397px\" \/>De acordo com a representa\u00e7\u00e3o gr\u00e1fica apresentada ao lado, o contradom\u00ednio da fun\u00e7\u00e3o $f$ \u00e9 $D{&#8216;_f} = \\left[ {1, + \\infty } \\right[$.\n<\/p>\n<\/li>\n<li>Como a fun\u00e7\u00e3o $f$ \u00e9 injetiva, ent\u00e3o admite inversa, sendo ${D_{{f^{ &#8211; 1}}}} = D{&#8216;_f} = \\left[ {1, + \\infty } \\right[$ e $D{&#8216;_{{f^{ &#8211; 1}}}} = {D_f} = \\left[ {0, + \\infty } \\right[$.\n<p>\\[\\begin{array}{*{20}{l}}<br \/>\n{y = 4{x^2} + 1}&amp; \\Leftrightarrow &amp;{{x^2} = \\frac{{y &#8211; 1}}{4}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{x =\u00a0 &#8211; \\sqrt {\\frac{{y &#8211; 1}}{4}} }&amp; \\vee &amp;{x = \\sqrt {\\frac{{y &#8211; 1}}{4}} }<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{x =\u00a0 &#8211; \\frac{{\\sqrt {y &#8211; 1} }}{2}}&amp; \\vee &amp;{x = \\frac{{\\sqrt {y &#8211; 1} }}{2}}<br \/>\n\\end{array}}<br \/>\n\\end{array}\\]<\/p>\n<p>Como $D{&#8216;_{{f^{ &#8211; 1}}}} = {D_f} = \\left[ {0, + \\infty } \\right[$, ent\u00e3o ${f^{ &#8211; 1}}\\left( x \\right) = \\frac{{\\sqrt {x &#8211; 1} }}{2}$.<\/p>\n<p>Portanto,\u00a0a fun\u00e7\u00e3o inversa de $f$ \u00e9: \\[\\begin{array}{*{20}{l}}<br \/>\n{{f^{ &#8211; 1}}:}&amp;{\\left[ {1, + \\infty } \\right[ \\to \\left[ {0, + \\infty } \\right[} \\\\<br \/>\n{}&amp;{x \\to \\frac{{\\sqrt {x &#8211; 1} }}{2}}<br \/>\n\\end{array}\\]<\/p>\n<p><strong>Nota<\/strong>: A\u00a0inversa da fun\u00e7\u00e3o de dom\u00ednio $\\left] { &#8211; \\infty ,0} \\right]$ e definida por $h\\left( x \\right) = 4{x^2} + 1$ \u00e9: \\[\\begin{array}{*{20}{l}}<br \/>\n{{h^{ &#8211; 1}}:}&amp;{\\left[ {1, + \\infty } \\right[ \\to \\left] { &#8211; \\infty ,0} \\right]} \\\\<br \/>\n{}&amp;{x \\to\u00a0 &#8211; \\frac{{\\sqrt {x &#8211; 1} }}{2}}<br \/>\n\\end{array}\\]<\/p>\n<\/li>\n<li>Conv\u00e9m notar que o dom\u00ednio da fun\u00e7\u00e3o $f \\circ g$\u00a0\u00e9: \\[{D_{f \\circ g}} = \\left\\{ {x \\in \\mathbb{R}:x \\in {D_g} \\wedge g\\left( x \\right) \\in {D_f}} \\right\\} = \\left\\{ {x \\in \\mathbb{R}:x \\in \\mathbb{R}_0^ +\u00a0 \\wedge \\sqrt {x + 6}\u00a0 \\in \\mathbb{R}_0^ + } \\right\\} = \\mathbb{R}_0^ + \\]\n<p>Resolvendo a condi\u00e7\u00e3o, vem: \\[\\begin{array}{*{20}{l}}<br \/>\n{f\\left( {g\\left( x \\right)} \\right) \\geqslant f\\left( x \\right)}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{4{{\\left( {\\sqrt {x + 6} } \\right)}^2} + 1 \\geqslant 4{x^2} + 1}&amp; \\wedge &amp;{x \\in \\mathbb{R}_0^ + }<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{4x + 24\\begin{array}{*{20}{l}}<br \/>\n{ + 1 \\geqslant 4{x^2} + 1}&amp; \\wedge &amp;{x \\in \\mathbb{R}_0^ + }<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{{x^2} &#8211; x &#8211; 6 \\leqslant 0}&amp; \\wedge &amp;{x \\in \\mathbb{R}_0^ + }<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{\\left( {x + 2} \\right)\\left( {x &#8211; 3} \\right) \\leqslant 0}&amp; \\wedge &amp;{x \\in \\mathbb{R}_0^ + }<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{x \\in \\left[ { &#8211; 2,3} \\right]}&amp; \\wedge &amp;{x \\in \\mathbb{R}_0^ + }<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x \\in \\left[ {0,3} \\right]}<br \/>\n\\end{array}\\]<\/p>\n<p>Situa\u00e7\u00e3o que tamb\u00e9m se pode verificar no gr\u00e1fico acima.<\/p>\n<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_11857' onClick='GTTabs_show(0,11857)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado A fun\u00e7\u00e3o $f$ tem dom\u00ednio $\\left[ {0, + \\infty } \\right[$ e \u00e9 definida por $f\\left( x \\right) = 4{x^2} + 1$. Esboce o gr\u00e1fico de $f$ e indique o contradom\u00ednio&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":20910,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[98,97,155],"tags":[422,156],"series":[],"class_list":["post-11857","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-11--ano","category-aplicando","category-funcao-inversa","tag-11-o-ano","tag-funcao-inversa-2"],"views":2031,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/04\/11V2Pag174-15_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/11857","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=11857"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/11857\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/20910"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=11857"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=11857"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=11857"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=11857"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}