{"id":6696,"date":"2011-04-05T01:53:48","date_gmt":"2011-04-05T00:53:48","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=6696"},"modified":"2021-12-26T16:09:00","modified_gmt":"2021-12-26T16:09:00","slug":"sendo-f-e-g-funcoes-reais-de-variavel-real-caracterize-fcirc-g-e-gcirc-f","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=6696","title":{"rendered":"Sendo $f$ e $g$ fun\u00e7\u00f5es reais de vari\u00e1vel real"},"content":{"rendered":"<p><ul id='GTTabs_ul_6696' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_6696' class='GTTabs_curr'><a  id=\"6696_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_6696' ><a  id=\"6696_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_6696'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Sendo $f$ e $g$ fun\u00e7\u00f5es reais de vari\u00e1vel real, caracterize $f\\circ g$ e $g\\circ f$ em cada um dos casos:<\/p>\n<ol>\n<li>$\\begin{matrix}<br \/>\nf(x)={{x}^{2}}+2x+1 &amp; e &amp; g(x)=3{{x}^{2}}+1\u00a0 \\\\<br \/>\n\\end{matrix}$<\/li>\n<li>$\\begin{matrix}<br \/>\nf(x)={{x}^{2}}+2x &amp; e &amp; g(x)=\\left| x \\right|+1\u00a0 \\\\<br \/>\n\\end{matrix}$<\/li>\n<li>$\\begin{matrix}<br \/>\nf(x)={{x}^{3}} &amp; e &amp; g(x)=\\frac{1}{x-3}\u00a0 \\\\<br \/>\n\\end{matrix}$<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_6696' onClick='GTTabs_show(1,6696)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_6696'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li>Ora, ${{D}_{f\\circ g}}=\\left\\{ x\\in \\mathbb{R}:x\\in {{D}_{g}}\\wedge g(x)\\in {{D}_{f}} \\right\\}=\\left\\{ x\\in \\mathbb{R}:x\\in \\mathbb{R}\\wedge (3{{x}^{2}}+1)\\in \\mathbb{R} \\right\\}=\\mathbb{R}$.<br \/>\nComo<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n(f\\circ g)(x) &amp; = &amp; f(g(x))\u00a0 \\\\<br \/>\n{} &amp; = &amp; f(3{{x}^{2}}+1)\u00a0 \\\\<br \/>\n{} &amp; = &amp; {{(3{{x}^{2}}+1)}^{2}}+2(3{{x}^{2}}+1)+1\u00a0 \\\\<br \/>\n{} &amp; = &amp; 9{{x}^{4}}+6{{x}^{2}}+1+6{{x}^{2}}+2+1\u00a0 \\\\<br \/>\n{} &amp; = &amp; 9{{x}^{4}}+12{{x}^{2}}+4\u00a0 \\\\<br \/>\n\\end{array}\\]<br \/>\nent\u00e3o, \\[\\begin{array}{*{35}{l}}<br \/>\nf\\circ g: &amp; \\mathbb{R}\\to \\mathbb{R}\u00a0 \\\\<br \/>\n{} &amp; x\\to 9{{x}^{4}}+12{{x}^{2}}+4\u00a0 \\\\<br \/>\n\\end{array}\\]<\/p>\n<p>Ora, ${{D}_{g\\circ f}}=\\left\\{ x\\in \\mathbb{R}:x\\in {{D}_{f}}\\wedge f(x)\\in {{D}_{g}} \\right\\}=\\left\\{ x\\in \\mathbb{R}:x\\in \\mathbb{R}\\wedge ({{x}^{2}}+2x+1)\\in \\mathbb{R} \\right\\}=\\mathbb{R}$.<br \/>\nComo<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n(g\\circ f)(x) &amp; = &amp; g(f(x))\u00a0 \\\\<br \/>\n{} &amp; = &amp; g({{x}^{2}}+2x+1)\u00a0 \\\\<br \/>\n{} &amp; = &amp; 3{{({{x}^{2}}+2x+1)}^{2}}+1\u00a0 \\\\<br \/>\n{} &amp; = &amp; 3({{x}^{2}}+2x+1)({{x}^{2}}+2x+1)+1\u00a0 \\\\<br \/>\n{} &amp; = &amp; 3({{x}^{4}}+2{{x}^{3}}+{{x}^{2}}+2{{x}^{3}}+4{{x}^{2}}+2x+{{x}^{2}}+2x+1)+1\u00a0 \\\\<br \/>\n{} &amp; = &amp; 3{{x}^{4}}+12{{x}^{3}}+18{{x}^{2}}+12x+4\u00a0 \\\\<br \/>\n\\end{array}\\]<br \/>\nent\u00e3o, \\[\\begin{array}{*{35}{l}}<br \/>\ng\\circ f: &amp; \\mathbb{R}\\to \\mathbb{R}\u00a0 \\\\<br \/>\n{} &amp; x\\to 3{{x}^{4}}+12{{x}^{3}}+18{{x}^{2}}+12x+4\u00a0 \\\\<br \/>\n\\end{array}\\]<\/p>\n<\/li>\n<li>Ora, ${{D}_{f\\circ g}}=\\left\\{ x\\in \\mathbb{R}:x\\in {{D}_{g}}\\wedge g(x)\\in {{D}_{f}} \\right\\}=\\left\\{ x\\in \\mathbb{R}:x\\in \\mathbb{R}\\wedge (\\left| x \\right|+1)\\in \\mathbb{R} \\right\\}=\\mathbb{R}$.<br \/>\nComo<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n(f\\circ g)(x) &amp; = &amp; f(g(x))\u00a0 \\\\<br \/>\n{} &amp; = &amp; f(\\left| x \\right|+1)\u00a0 \\\\<br \/>\n{} &amp; = &amp; {{(\\left| x \\right|+1)}^{2}}+2(\\left| x \\right|+1)\u00a0 \\\\<br \/>\n{} &amp; = &amp; {{x}^{2}}+2\\left| x \\right|+1+2\\left| x \\right|+2\u00a0 \\\\<br \/>\n{} &amp; = &amp; {{x}^{2}}+4\\left| x \\right|+3\u00a0 \\\\<br \/>\n\\end{array}\\]<br \/>\nent\u00e3o, \\[\\begin{array}{*{35}{l}}<br \/>\nf\\circ g: &amp; \\mathbb{R}\\to \\mathbb{R}\u00a0 \\\\<br \/>\n{} &amp; x\\to {{x}^{2}}+4\\left| x \\right|+3\u00a0 \\\\<br \/>\n\\end{array}\\]<\/p>\n<p>Ora, ${{D}_{g\\circ f}}=\\left\\{ x\\in \\mathbb{R}:x\\in {{D}_{f}}\\wedge f(x)\\in {{D}_{g}} \\right\\}=\\left\\{ x\\in \\mathbb{R}:x\\in \\mathbb{R}\\wedge ({{x}^{2}}+2x)\\in \\mathbb{R} \\right\\}=\\mathbb{R}$.<br \/>\nComo<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n(g\\circ f)(x) &amp; = &amp; g(f(x))\u00a0 \\\\<br \/>\n{} &amp; = &amp; g({{x}^{2}}+2x)\u00a0 \\\\<br \/>\n{} &amp; = &amp; \\left| {{x}^{2}}+2x \\right|+1\u00a0 \\\\<br \/>\n\\end{array}\\]<br \/>\nent\u00e3o, \\[\\begin{array}{*{35}{l}}<br \/>\ng\\circ f: &amp; \\mathbb{R}\\to \\mathbb{R}\u00a0 \\\\<br \/>\n{} &amp; x\\to \\left| {{x}^{2}}+2x \\right|+1\u00a0 \\\\<br \/>\n\\end{array}\\]<\/p>\n<\/li>\n<li>Ora, ${{D}_{f\\circ g}}=\\left\\{ x\\in \\mathbb{R}:x\\in {{D}_{g}}\\wedge g(x)\\in {{D}_{f}} \\right\\}=\\left\\{ x\\in \\mathbb{R}:x\\in \\mathbb{R}\\backslash \\left\\{ 3 \\right\\}\\wedge (\\frac{1}{x-3})\\in \\mathbb{R} \\right\\}=\\mathbb{R}\\backslash \\left\\{ 3 \\right\\}$.<br \/>\nComo<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n(f\\circ g)(x) &amp; = &amp; f(g(x))\u00a0 \\\\<br \/>\n{} &amp; = &amp; f(\\frac{1}{x-3})\u00a0 \\\\<br \/>\n{} &amp; = &amp; {{(\\frac{1}{x-3})}^{3}}\u00a0 \\\\<br \/>\n{} &amp; = &amp; \\frac{1}{{{(x-3)}^{3}}}\u00a0 \\\\<br \/>\n\\end{array}\\]<br \/>\nent\u00e3o, \\[\\begin{array}{*{35}{l}}<br \/>\nf\\circ g: &amp; \\mathbb{R}\\backslash \\left\\{ 3 \\right\\}\\to \\mathbb{R}\u00a0 \\\\<br \/>\n{} &amp; x\\to \\frac{1}{{{(x-3)}^{3}}}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/p>\n<p>Ora, ${{D}_{g\\circ f}}=\\left\\{ x\\in \\mathbb{R}:x\\in {{D}_{f}}\\wedge f(x)\\in {{D}_{g}} \\right\\}=\\left\\{ x\\in \\mathbb{R}:x\\in \\mathbb{R}\\wedge ({{x}^{3}})\\in \\mathbb{R}\\backslash \\left\\{ 3 \\right\\} \\right\\}=\\mathbb{R}\\backslash \\left\\{ \\sqrt[3]{3} \\right\\}$.<br \/>\nComo<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n(g\\circ f)(x) &amp; = &amp; g(f(x))\u00a0 \\\\<br \/>\n{} &amp; = &amp; g({{x}^{3}})\u00a0 \\\\<br \/>\n{} &amp; = &amp; \\frac{1}{{{x}^{3}}-3}\u00a0 \\\\<br \/>\n\\end{array}\\]<br \/>\nent\u00e3o, \\[\\begin{array}{*{35}{l}}<br \/>\ng\\circ f: &amp; \\mathbb{R}\\backslash \\left\\{ \\sqrt[3]{3} \\right\\}\\to \\mathbb{R}\u00a0 \\\\<br \/>\n{} &amp; x\\to \\frac{1}{{{x}^{3}}-3}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/p>\n<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_6696' onClick='GTTabs_show(0,6696)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Sendo $f$ e $g$ fun\u00e7\u00f5es reais de vari\u00e1vel real, caracterize $f\\circ g$ e $g\\circ f$ em cada um dos casos: $\\begin{matrix} f(x)={{x}^{2}}+2x+1 &amp; e &amp; g(x)=3{{x}^{2}}+1\u00a0 \\\\ \\end{matrix}$ $\\begin{matrix} f(x)={{x}^{2}}+2x &amp;&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":19179,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[98,97,153],"tags":[154],"series":[],"class_list":["post-6696","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-11--ano","category-aplicando","category-funcao-composta","tag-funcao-composta-2"],"views":1963,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2021\/12\/Mat70.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/6696","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=6696"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/6696\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/19179"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=6696"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=6696"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=6696"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=6696"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}