{"id":11831,"date":"2014-02-27T23:31:38","date_gmt":"2014-02-27T23:31:38","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=11831"},"modified":"2021-12-26T15:53:32","modified_gmt":"2021-12-26T15:53:32","slug":"considere-as-funcoes-f-e-g-2","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=11831","title":{"rendered":"Considere as fun\u00e7\u00f5es $f$ e $g$"},"content":{"rendered":"<p><ul id='GTTabs_ul_11831' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_11831' class='GTTabs_curr'><a  id=\"11831_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_11831' ><a  id=\"11831_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_11831'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Considere as fun\u00e7\u00f5es $f$ e $g$ definidas por: \\[\\begin{array}{*{20}{c}}<br \/>\n{f\\left( x \\right) = \\frac{2}{{x &#8211; 1}}}&amp;{\\text{e}}&amp;{g\\left( x \\right) = \\frac{1}{{x + 3}}}<br \/>\n\\end{array}\\]<\/p>\n<ol>\n<li>Determine o dom\u00ednio de cada uma delas.<\/li>\n<li>Caracterize as fun\u00e7\u00f5es $f \\circ g$, $g \\circ f$ e $f \\circ f$.<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_11831' onClick='GTTabs_show(1,11831)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_11831'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<blockquote>\n<p>\\[\\begin{array}{*{20}{c}}<br \/>\n{f\\left( x \\right) = \\frac{2}{{x &#8211; 1}}}&amp;{\\text{e}}&amp;{g\\left( x \\right) = \\frac{1}{{x + 3}}}<br \/>\n\\end{array}\\]<\/p>\n<\/blockquote>\n<ol>\n<li>${D_f} = \\left\\{ {x \\in \\mathbb{R}:x &#8211; 1 \\ne 0} \\right\\} = \\mathbb{R}\\backslash \\left\\{ 1 \\right\\}$.\n<p>${D_g} = \\left\\{ {x \\in \\mathbb{R}:x + 3 \\ne 0} \\right\\} = \\mathbb{R}\\backslash \\left\\{ { &#8211; 3} \\right\\}$.<\/p>\n<\/li>\n<li>\\[{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}\\backslash \\left\\{ { &#8211; 3} \\right\\} \\wedge \\left( {\\frac{1}{{x + 3}}} \\right) \\in \\mathbb{R}\\backslash \\left\\{ { &#8211; 3, &#8211; 2} \\right\\}} \\right\\} = \\mathbb{R}\\backslash \\left\\{ { &#8211; 3, &#8211; 2} \\right\\}\\]<br \/>\n\\[{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}\\backslash \\left\\{ { &#8211; 3} \\right\\} \\wedge \\left( {\\frac{1}{{x + 3}}} \\right) \\in \\mathbb{R}\\backslash \\left\\{ 1 \\right\\}} \\right\\} = \\mathbb{R}\\backslash \\left\\{ { &#8211; 3, &#8211; 2} \\right\\}\\]<br \/>\n\\[\\begin{array}{*{20}{c}}<br \/>\n{f \\circ g:}&amp;{\\mathbb{R}\\backslash \\left\\{ { &#8211; 3, &#8211; 2} \\right\\} \\to \\mathbb{R}} \\\\<br \/>\n{}&amp;{x \\to\u00a0 &#8211; \\frac{{2x + 6}}{{x + 2}}}<br \/>\n\\end{array}\\]<\/p>\n<p>\\[{D_{g \\circ f}} = \\left\\{ {x \\in \\mathbb{R}:x \\in {D_f} \\wedge f\\left( x \\right) \\in {D_g}} \\right\\} = \\left\\{ {x \\in \\mathbb{R}:x \\in \\mathbb{R}\\backslash \\left\\{ 1 \\right\\} \\wedge \\left( {\\frac{2}{{x &#8211; 1}}} \\right) \\in \\mathbb{R}\\backslash \\left\\{ { &#8211; 3} \\right\\}} \\right\\} = \\mathbb{R}\\backslash \\left\\{ {\\frac{1}{3},1} \\right\\}\\]<br \/>\n\\[\\left( {g \\circ f} \\right)\\left( x \\right) = g\\left( {f\\left( x \\right)} \\right) = g\\left( {\\frac{2}{{x &#8211; 1}}} \\right) = \\frac{1}{{\\frac{2}{{x &#8211; 1}} + 3}} = \\frac{1}{{\\frac{{2 + 3x &#8211; 3}}{{x &#8211; 1}}}} = \\frac{{x &#8211; 1}}{{3x &#8211; 1}},\\,\\forall x \\in {D_{g \\circ f}}\\]<br \/>\n\\[\\begin{array}{*{20}{c}}<br \/>\n{g \\circ f:}&amp;{\\mathbb{R}\\backslash \\left\\{ {\\frac{1}{3},1} \\right\\} \\to \\mathbb{R}} \\\\<br \/>\n{}&amp;{x \\to \\frac{{x &#8211; 1}}{{3x &#8211; 1}}}<br \/>\n\\end{array}\\]<\/p>\n<p>\\[{D_{f \\circ f}} = \\left\\{ {x \\in \\mathbb{R}:x \\in {D_f} \\wedge f\\left( x \\right) \\in {D_f}} \\right\\} = \\left\\{ {x \\in \\mathbb{R}:x \\in \\mathbb{R}\\backslash \\left\\{ 1 \\right\\} \\wedge \\left( {\\frac{2}{{x &#8211; 1}}} \\right) \\in \\mathbb{R}\\backslash \\left\\{ 1 \\right\\}} \\right\\} = \\mathbb{R}\\backslash \\left\\{ {1,3} \\right\\}\\]<br \/>\n\\[\\left( {f \\circ f} \\right)\\left( x \\right) = f\\left( {f\\left( x \\right)} \\right) = f\\left( {\\frac{2}{{x &#8211; 1}}} \\right) = \\frac{2}{{\\frac{2}{{x &#8211; 1}} &#8211; 1}} = \\frac{2}{{\\frac{{2 &#8211; x + 1}}{{x &#8211; 1}}}} = \\frac{{2x &#8211; 2}}{{3 &#8211; x}},\\,\\forall x \\in {D_{f \\circ f}}\\]<br \/>\n\\[\\begin{array}{*{20}{c}}<br \/>\n{f \\circ f:}&amp;{\\mathbb{R}\\backslash \\left\\{ {1,3} \\right\\} \\to \\mathbb{R}} \\\\<br \/>\n{}&amp;{x \\to \\frac{{2x &#8211; 2}}{{3 &#8211; x}}}<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_11831' onClick='GTTabs_show(0,11831)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Considere as fun\u00e7\u00f5es $f$ e $g$ definidas por: \\[\\begin{array}{*{20}{c}} {f\\left( x \\right) = \\frac{2}{{x &#8211; 1}}}&amp;{\\text{e}}&amp;{g\\left( x \\right) = \\frac{1}{{x + 3}}} \\end{array}\\] Determine o dom\u00ednio de cada uma delas. Caracterize&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":19189,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[98,97,153],"tags":[422,154],"series":[],"class_list":["post-11831","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-11--ano","category-aplicando","category-funcao-composta","tag-11-o-ano","tag-funcao-composta-2"],"views":3259,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2021\/12\/Mat75.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/11831","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=11831"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/11831\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/19189"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=11831"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=11831"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=11831"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=11831"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}