\n\\[\\begin{array}{*{20}{l}}
\n{\\begin{array}{*{20}{l}}
\n{f:}&{\\mathbb{R}\\backslash \\left\\{ 0 \\right\\} \\to \\mathbb{R}} \\\\
\n{}&{x \\to \\frac{1}{{{x^2}}}}
\n\\end{array}}&{}&{}&{\\begin{array}{*{20}{l}}
\n{g:}&{\\mathbb{R}\\backslash \\left\\{ { – 1} \\right\\} \\to \\mathbb{R}} \\\\
\n{}&{x \\to x + 1}
\n\\end{array}}&{}&{}&{\\begin{array}{*{20}{l}}
\n{h:}&{\\mathbb{R}\\backslash \\left\\{ {0,1} \\right\\} \\to \\mathbb{R}} \\\\
\n{}&{x \\to \\frac{1}{{{x^2} – x}}}
\n\\end{array}}
\n\\end{array}\\]<\/p>\n<\/blockquote>\n
\n\\[{f + g}\\]<\/span><\/p>\n<\/p>\n\\[{D_{f + g}} = {D_f} \\cap {D_g} = \\mathbb{R}\\backslash \\left\\{ 0 \\right\\} \\cap \\mathbb{R}\\backslash \\left\\{ { – 1} \\right\\} = \\mathbb{R}\\backslash \\left\\{ { – 1,0} \\right\\}\\]<\/p>\n
\\[\\left( {f + g} \\right)\\left( x \\right) = f\\left( x \\right) + g\\left( x \\right) = \\frac{1}{{{x^2}}} + x + 1 = \\frac{{{x^3} + {x^2} + 1}}{{{x^2}}},\\forall x \\in {D_{f + g}}\\]<\/p>\n
\\[\\begin{array}{*{20}{l}}
\n{f + g:}&{\\mathbb{R}\\backslash \\left\\{ { – 1,0} \\right\\} \\to \\mathbb{R}} \\\\
\n{}&{x \\to \\frac{{{x^3} + {x^2} + 1}}{{{x^2}}}}
\n\\end{array}\\]<\/p>\n
\n\\[{f \\times g}\\]<\/span><\/p>\n<\/p>\n\\[{D_{f \\times g}} = {D_f} \\cap {D_g} = \\mathbb{R}\\backslash \\left\\{ 0 \\right\\} \\cap \\mathbb{R}\\backslash \\left\\{ { – 1} \\right\\} = \\mathbb{R}\\backslash \\left\\{ { – 1,0} \\right\\}\\]<\/p>\n
\\[\\left( {f \\times g} \\right)\\left( x \\right) = f\\left( x \\right) \\times g\\left( x \\right) = \\frac{1}{{{x^2}}} \\times \\left( {x + 1} \\right) = \\frac{{x + 1}}{{{x^2}}},\\forall x \\in {D_{f \\times g}}\\]<\/p>\n
\\[\\begin{array}{*{20}{l}}
\n{f \\times g:}&{\\mathbb{R}\\backslash \\left\\{ { – 1,0} \\right\\} \\to \\mathbb{R}} \\\\
\n{}&{x \\to \\frac{{x + 1}}{{{x^2}}}}
\n\\end{array}\\]<\/p>\n
\n\\[{\\frac{f}{g}}\\]<\/span><\/p>\n<\/p>\n\\[{D_{\\frac{f}{g}}} = {D_f} \\cap {D_g} \\cap \\left\\{ {x \\in \\mathbb{R}:g\\left( x \\right) \\ne 0} \\right\\} = \\mathbb{R}\\backslash \\left\\{ 0 \\right\\} \\cap \\mathbb{R}\\backslash \\left\\{ { – 1} \\right\\} \\cap \\mathbb{R}\\backslash \\left\\{ { – 1} \\right\\} = \\mathbb{R}\\backslash \\left\\{ { – 1,0} \\right\\}\\]<\/p>\n
\\[\\left( {\\frac{f}{g}} \\right)\\left( x \\right) = \\frac{{f\\left( x \\right)}}{{g\\left( x \\right)}} = \\frac{{\\frac{1}{{{x^2}}}}}{{x + 1}} = \\frac{1}{{{x^3} + {x^2}}},\\forall x \\in {D_{\\frac{f}{g}}}\\]<\/p>\n
\\[\\begin{array}{*{20}{l}}
\n{\\frac{f}{g}:}&{\\mathbb{R}\\backslash \\left\\{ { – 1,0} \\right\\} \\to \\mathbb{R}} \\\\
\n{}&{x \\to \\frac{1}{{{x^3} + {x^2}}}}
\n\\end{array}\\]<\/p>\n<\/p>\n
\n\\[{h – g}\\]<\/span><\/p>\n<\/p>\n\\[{D_{h – g}} = {D_h} \\cap {D_g} = \\mathbb{R}\\backslash \\left\\{ {0,1} \\right\\} \\cap \\mathbb{R}\\backslash \\left\\{ { – 1} \\right\\} = \\mathbb{R}\\backslash \\left\\{ { – 1,0,1} \\right\\}\\]<\/p>\n
\\[\\left( {h – g} \\right)\\left( x \\right) = h\\left( x \\right) – g\\left( x \\right) = \\frac{1}{{{x^2} – x}} – \\left( {x + 1} \\right) = \\frac{{ – {x^3} – {x^2} + {x^2} + x + 1}}{{{x^2} – x}} = \\frac{{ – {x^3} + x + 1}}{{{x^2} – x}},\\forall x \\in {D_{\\frac{f}{g}}}\\]<\/p>\n
\\[\\begin{array}{*{20}{l}}
\n{h – g:}&{\\mathbb{R}\\backslash \\left\\{ { – 1,0,1} \\right\\} \\to \\mathbb{R}} \\\\
\n{}&{x \\to \\frac{{ – {x^3} + x + 1}}{{{x^2} – x}}}
\n\\end{array}\\]<\/p>\n<\/p>\n
\n\\[\\frac{f}{h}\\]<\/span><\/p>\n<\/p>\n\\[{D_{\\frac{f}{h}}} = {D_f} \\cap {D_h} \\cap \\left\\{ {x \\in \\mathbb{R}:h\\left( x \\right) \\ne 0} \\right\\} = \\mathbb{R}\\backslash \\left\\{ 0 \\right\\} \\cap \\mathbb{R}\\backslash \\left\\{ {0,1} \\right\\} \\cap \\mathbb{R}\\backslash \\left\\{ {0,1} \\right\\} = \\mathbb{R}\\backslash \\left\\{ {0,1} \\right\\}\\]<\/p>\n
\\[\\left( {\\frac{f}{h}} \\right)\\left( x \\right) = \\frac{{f\\left( x \\right)}}{{h\\left( x \\right)}} = \\frac{{\\frac{1}{{{x^2}}}}}{{\\frac{1}{{{x^2} – x}}}} = \\frac{{{x^2} – x}}{{{x^2}}} = \\frac{{x\\left( {x – 1} \\right)}}{{{x^2}}} = \\frac{{x – 1}}{x},\\forall x \\in {D_{\\frac{f}{g}}}\\]<\/p>\n
\\[\\begin{array}{*{20}{l}}
\n{\\frac{f}{h}:}&{\\mathbb{R}\\backslash \\left\\{ {0,1} \\right\\} \\to \\mathbb{R}} \\\\
\n{}&{x \\to \\frac{{x – 1}}{x}}
\n\\end{array}\\]<\/p>\n<\/p>\n
\n\\[f \\circ g\\]<\/span><\/p>\n<\/p>\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\\{ { – 1} \\right\\} \\wedge \\left( {x + 1} \\right) \\in \\mathbb{R}\\backslash \\left\\{ 0 \\right\\}} \\right\\} = \\left\\{ {x \\in \\mathbb{R}:x \\in \\mathbb{R}\\backslash \\left\\{ { – 1} \\right\\} \\wedge x \\in \\mathbb{R}\\backslash \\left\\{ { – 1} \\right\\}} \\right\\} = \\mathbb{R}\\backslash \\left\\{ { – 1} \\right\\}\\]<\/p>\n
\\[\\left( {f \\circ g} \\right)\\left( x \\right) = f\\left( {g\\left( x \\right)} \\right) = f\\left( {x + 1} \\right) = \\frac{1}{{{{\\left( {x + 1} \\right)}^2}}},\\forall x \\in {D_{f \\circ g}}\\]<\/p>\n
\\[\\begin{array}{*{20}{l}}
\n{f \\circ g:}&{\\mathbb{R}\\backslash \\left\\{ { – 1} \\right\\} \\to \\mathbb{R}} \\\\
\n{}&{x \\to \\frac{1}{{{{\\left( {x + 1} \\right)}^2}}}}
\n\\end{array}\\]<\/p>\n<\/p>\n
\n\\[g \\circ f\\]<\/span><\/p>\n<\/p>\n\\[{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\\{ 0 \\right\\} \\wedge \\left( {\\frac{1}{{{x^2}}}} \\right) \\in \\mathbb{R}\\backslash \\left\\{ { – 1} \\right\\}} \\right\\} = \\left\\{ {x \\in \\mathbb{R}:x \\in \\mathbb{R}\\backslash \\left\\{ 0 \\right\\} \\wedge x \\in \\mathbb{R}\\backslash \\left\\{ 0 \\right\\}} \\right\\} = \\mathbb{R}\\backslash \\left\\{ 0 \\right\\}\\]<\/p>\n
\\[\\left( {g \\circ f} \\right)\\left( x \\right) = g\\left( {f\\left( x \\right)} \\right) = g\\left( {\\frac{1}{{{x^2}}}} \\right) = \\frac{1}{{{x^2}}} + 1 = \\frac{{{x^2} + 1}}{{{x^2}}},\\forall x \\in {D_{f \\circ g}}\\]<\/p>\n
\\[\\begin{array}{*{20}{l}}
\n{g \\circ f:}&{\\mathbb{R}\\backslash \\left\\{ 0 \\right\\} \\to \\mathbb{R}} \\\\
\n{}&{x \\to \\frac{{{x^2} + 1}}{{{x^2}}}}
\n\\end{array}\\]<\/p>\n
<< Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"Enunciado Resolu\u00e7\u00e3o Enunciado Considere as fun\u00e7\u00f5es definidas por: \u00a0\\[\\begin{array}{*{20}{l}} {\\begin{array}{*{20}{l}} {f:}&{\\mathbb{R}\\backslash \\left\\{ 0 \\right\\} \\to \\mathbb{R}} \\\\ {}&{x \\to \\frac{1}{{{x^2}}}} \\end{array}}&{}&{}&{\\begin{array}{*{20}{l}} {g:}&{\\mathbb{R}\\backslash \\left\\{ { – 1} \\right\\} \\to \\mathbb{R}} \\\\ {}&{x \\to x +...<\/p>\n","protected":false},"author":1,"featured_media":19175,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[98,97,153,147],"tags":[422,154,363,366,364,365],"series":[],"class_list":["post-11826","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-11--ano","category-aplicando","category-funcao-composta","category-operacoes-com-funcoes","tag-11-o-ano","tag-funcao-composta-2","tag-funcao-diferenca","tag-funcao-produto","tag-funcao-quociente","tag-funcao-soma"],"views":3702,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2021\/12\/Mat66.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/11826","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=11826"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/11826\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/19175"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=11826"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=11826"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=11826"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=11826"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}