{"id":205356,"date":"2023-11-10T05:01:36","date_gmt":"2023-11-10T08:01:36","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=205356"},"modified":"2023-11-10T05:02:34","modified_gmt":"2023-11-10T08:02:34","slug":"a-um-pouco-mais-de-composicao-de-funcoes","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/a-um-pouco-mais-de-composicao-de-funcoes\/","title":{"rendered":"(A) Um pouco mais de composi\u00e7\u00e3o de fun\u00e7\u00f5es"},"content":{"rendered":"<p class=\"ti3\"  style=\"text-align:center;\">Clique no bot\u00e3o abaixo para visualizar a Sala.<\/p>\n<div style=\"text-align:center;margin-top:10px;\">\n<input class=\"bo1 c1 cb13\" style=\"width:150px;\" value=\"Sala\" title=\"Clique aqui\" type=\"button\" onclick=    \"if(this.parentNode.getElementsByTagName('div')[0].style.display != ''){this.parentNode.getElementsByTagName('div')[0].style.display = '';this.value = 'Ocultar';} else{this.parentNode.getElementsByTagName('div')[0].style.display = 'none'; this.value = 'Sala';}\" \/>  <\/p>\n<div style=\"display: none;\">\n<div class=\"gr1 cbm1\" style=\"color: #1e1e1e; font-size: 15px; line-height: 30px;\">\n<div class=\"nome cm\" style=\"font-size: 40px; line-height: 60px\"><strong>Um pouco mais de composi\u00e7\u00e3o de fun\u00e7\u00f5es<\/strong><\/div>\n<hr style=\"height:2px; width:100%;background-color:#8B5A2B;   margin-top: 8px; margin-bottom: 8px; border: 0 none\"\/>\n<hr style=\"height:2px; width:100%;background-color:#8B5A2B;   margin-top: 8px; margin-bottom: 20px; border: 0 none\"\/>\nNesta Sala vamos fazer alguns exemplos mais envolvendo composi\u00e7\u00e3o de fun\u00e7\u00f5es.<\/p>\n<p style=\"margin:10px;\">\n<div class=\"gr1 cbm5\" style=\"color: #1e1e1e; padding:20px; width:95%;font-size: 15px; line-height: 30px\">\n<strong><span style=\"color: #8B5A2B;\">Exemplo 1:<\/span><\/strong> Considere as seguintes fun\u00e7\u00f5es: <\/p>\n<p style=\"margin:10px;\">\n[tex]\\qquad \\begin{eqnarray*}<br \/>\nf:\\mathbb{R} &#038;\\rightarrow &#038; \\mathbb{R}\\qquad \\qquad g:&#038; \\mathbb{R} \\rightarrow &#038;  \\mathbb{R}\\\\<br \/>\nx&#038;\\mapsto &#038; x^2  &#038; x\\mapsto &#038; x+2<br \/>\n\\end{eqnarray*}[\/tex]<\/p>\n<p>\u00c9 poss\u00edvel definir [tex]f\\circ g~[\/tex] e [tex]~g\\circ f[\/tex]?<\/p>\n<p style=\"margin:10px;\">\n<p><span style=\"color:#98622F;\">&#9658;<\/span> Observe inicialmente que [tex]x^2 \\geqslant 0[\/tex] para todo [tex]x \\in \\mathbb{R}[\/tex]; portanto, [tex]Im(f)=\\{y \\in \\mathbb{R}~|~y \\geqslant 0\\}. [\/tex]<br \/>\nCom isso, [tex]Im(f)\\subset D(g)=\\mathbb{R}[\/tex] e a fun\u00e7\u00e3o [tex]~g\\circ f[\/tex] est\u00e1 definida.<br \/>\nAgora, note que:<br \/>\n[tex]\\qquad \\left(g\\circ f\\right)(x)=g\\left(f(x)\\right)=g\\left(x^2\\right)=x^2+2.[\/tex]<br \/>\nLogo, temos:<br \/>\n[tex]\\qquad g\\circ f:\\mathbb{R}\\rightarrow  \\mathbb{R}\\\\<br \/>\n\\qquad \\left(g\\circ f\\right)(x)=x^2+2.<br \/>\n[\/tex]<\/p>\n<p style=\"margin:10px;\">\n<p><span style=\"color:#98622F;\">&#9658;<\/span> Observe, agora, que [tex]x+2 \\in \\mathbb{R}[\/tex] para todo [tex]x \\in \\mathbb{R}[\/tex]; portanto, [tex]Im(g)\\subset\\mathbb{R}. [\/tex]<br \/>\nCom isso, [tex]Im(g)\\subset D(f)=\\mathbb{R}[\/tex] e a fun\u00e7\u00e3o [tex]~f\\circ g[\/tex] tamb\u00e9m est\u00e1 definida:<br \/>\n[tex]\\qquad \\left(f\\circ g\\right)(x)=f\\left(g(x)\\right)=f\\left(x+2\\right)=\\left(x+2\\right)^2.[\/tex]<br \/>\nLogo, temos:<br \/>\n[tex]\\qquad f\\circ g:\\mathbb{R}\\rightarrow  \\mathbb{R}\\\\<br \/>\n\\qquad  \\left(f\\circ g\\right)=x^2+2x+4.<br \/>\n[\/tex]<\/p>\n<p style=\"margin:10px;\">\n<p>Observe que, embora as duas compostas estejam definidas, elas s\u00e3o distintas!\n<\/p><\/div>\n<p style=\"margin:20px;\">\n<div class=\"gr1 cbm5\" style=\"color: #1e1e1e; padding:20px; width:95%;font-size: 15px;line-height: 30px;\">\n<strong><span style=\"color: #8B5A2B;\">Exemplo 2:<\/span><\/strong> Considere as fun\u00e7\u00f5es de [tex]\\mathbb{R}[\/tex] em [tex]\\mathbb{R}[\/tex] definidas por:<br \/>\n[tex]\\qquad f(x)=2x-5\\quad[\/tex] e [tex]\\quad g(x)=x+20[\/tex], determine o valor de:<\/p>\n<p style=\"margin:10px;\">\n<p><strong>a)<\/strong> [tex]\\textcolor{#BF7B3B}{ \\left(f\\circ g\\right)(2)}=f(g(2))=f(22)=2\\times 22-5=44-5=39.[\/tex]<br \/>\n<strong>b)<\/strong> [tex]\\textcolor{#BF7B3B}{ \\left(g\\circ f\\right)(0)}=g(f(0))=g(0-5)=g(-5)=-5+20=15.[\/tex]<br \/>\n<strong>c)<\/strong> [tex]\\textcolor{#BF7B3B}{ \\left(f\\circ f\\right)(1)}=f(f(1))=f(-3)=-11.[\/tex]<br \/>\n<strong>d)<\/strong> [tex]\\textcolor{#BF7B3B}{ \\left(g\\circ g\\right)(5)}=g(g(5))=g(25)=45.[\/tex]<br \/>\n<strong>e)<\/strong> [tex]\\textcolor{#BF7B3B}{\\left(g\\circ g\\circ g\\right)(3)}=g(g(g(3)))=g(g(23))=g(43)=63.[\/tex]\n<\/div>\n<p style=\"margin:20px;\">\n<div class=\"gr1 cbm5\" style=\"color: #1e1e1e; padding:20px; width:95%;font-size: 15px;line-height: 30px;\">\n<strong><span style=\"color: #8B5A2B;\">Exemplo 3:<\/span><\/strong> Um corpo \u00e9 lan\u00e7ado verticalmente para cima a partir do solo, com velocidade inicial de [tex]30\\text{ m\/s}.[\/tex]<br \/>\nDesprezando-se a resist\u00eancia do ar e considerando-se que a acelera\u00e7\u00e3o da gravidade \u00e9 de [tex]g=10 \\text{ m\/s}^2[\/tex], ent\u00e3o, a cada instante [tex]t[\/tex] dado em segundos, sua altura em rela\u00e7\u00e3o \u00e0 superf\u00edcie da terra \u00e9 dada em metros pela fun\u00e7\u00e3o [tex]h[\/tex] definida por<br \/>\n[tex]\\qquad h(t)=-\\dfrac{g}{2}\\,t^2+v_0t\\\\<br \/>\n\\qquad \\boxed{h(t)=-5t^2+30t}.[\/tex]<br \/>\nE se fosse necess\u00e1rio determinarmos a altura do corpo a partir do tempo dado em minutos?<\/p>\n<p style=\"margin:10px;\">\n<p><span style=\"color:#98622F;\">&#9658;<\/span> Para converter minutos em segundos, podemos utilizar a fun\u00e7\u00e3o [tex]c[\/tex] definida por<br \/>\n[tex] \\qquad c(t)=60t[\/tex]<br \/>\ncom [tex]t[\/tex] dado em minutos.<br \/>\nDessa forma, a fun\u00e7\u00e3o composta [tex]h\\circ c[\/tex] nos d\u00e1 a altura do corpo a partir do tempo [tex]t[\/tex] dado em minutos. Se denotarmos essa composta por [tex]k[\/tex], segue que:<br \/>\n[tex]\\quad k(t)=\\left(h\\circ c\\right)(t)=h(c(t))\\\\<br \/>\n\\quad k(t)=h(60t)=-5\\left(60t\\right)^2+30(60t)\\\\<br \/>\n\\quad k(t)=18000t^2+1800t.[\/tex]<br \/>\nAssim, a altura em metros do corpo em rela\u00e7\u00e3o \u00e0 superf\u00edcie da terra a cada instante [tex]t[\/tex] dado em minutos \u00e9 dada pela fun\u00e7\u00e3o [tex]k[\/tex] definida por [tex]\\boxed{k(t)=18000t^2+1800t}.[\/tex]\n<\/div>\n<p style=\"margin:20px;\">\n<div class=\"gr1 cbm5\" style=\"color: #1e1e1e; padding:20px; width:95%;font-size: 15px;line-height: 30px;\">\n<strong><span style=\"color: #8B5A2B;\">Exemplo 4:<\/span><\/strong> Considere as fun\u00e7\u00f5es [tex]f[\/tex], [tex]g[\/tex], [tex]h[\/tex] e [tex]t[\/tex] assim definidas:<br \/>\n[tex]~~ f(x)=\\dfrac{2-x}{x}[\/tex] ; [tex]g(x)=\\sqrt[3]{x+1}-2[\/tex] ; [tex]h(x)=(x+2)^3-1[\/tex] ; [tex]t(x)=\\dfrac{2}{x+1}.[\/tex]<br \/>\nQual a lei de forma\u00e7\u00e3o da composta [tex] f \\circ g \\circ h \\circ t[\/tex]?<\/p>\n<p style=\"margin:10px;\">\n<p><span style=\"color:#98622F;\">&#9658;<\/span> Observe o desenvolvimento a seguir:<br \/>\n[tex]\\qquad \\begin{align*} \\left(f \\circ g \\circ h \\circ t\\right)(x)&#038;=f(g(h(t(x))))\\\\<br \/>\n&#038;=f\\left(g\\left(h\\left(\\dfrac{2}{x+1}\\right)\\right)\\right)\\\\<br \/>\n&#038;=f\\left(g\\left(\\left(\\dfrac{2}{x+1}+2\\right)^3-1\\right)\\right)\\\\<br \/>\n&#038;=f\\left(\\sqrt[3]{\\left(\\left(\\dfrac{2}{x+1}+2\\right)^3-1\\right)+1~}-2\\right)\\\\<br \/>\n&#038;=f\\left(\\sqrt[3]{\\left(\\dfrac{2}{x+1}+2\\right)^3}-2\\right)\\\\<br \/>\n&#038;=f\\left(\\left(\\dfrac{2}{x+1}+2\\right)-2\\right)\\\\<br \/>\n&#038;=f\\left(\\dfrac{2}{x+1}\\right)\\\\<br \/>\n&#038;=\\dfrac{2-\\dfrac{2}{x+1}}{\\dfrac{2}{x+1}}=\\dfrac{\\dfrac{2x+2-2}{x+1}}{\\dfrac{2}{x+1}}=\\dfrac{\\dfrac{2x}{\\cancel{x+1}}}{\\dfrac{2}{\\cancel{x+1}}}\\\\<br \/>\n&#038;=\\dfrac{2x}{2}=x<br \/>\n\\end{align*} [\/tex]<br \/>\nPortanto, [tex]\\left(f \\circ g \\circ h \\circ t\\right)(x)=x.[\/tex]\n<\/div>\n<p style=\"margin:20px;\">\n<div class=\"gr1 cbm5\" style=\"color: #1e1e1e; padding:20px; width:95%;font-size: 15px;line-height: 30px;\">\n<strong><span style=\"color: #8B5A2B;\">Exemplo 5:<\/span><\/strong> Em um determinado rio, peixes grandes de uma certa esp\u00e9cie se alimentam de pequenos peixes e esses se alimentam de pl\u00e2nctons.<br \/>\nSuponhamos que [tex]f[\/tex] seja a fun\u00e7\u00e3o que nos d\u00e1 o tamanho da popula\u00e7\u00e3o desses peixes grandes, a partir do n\u00famero [tex]n[\/tex] de peixes pequenos presentes no rio, e  [tex]g[\/tex] seja a fun\u00e7\u00e3o que expressa o n\u00famero desses peixes pequenos, a partir da quantidade [tex]x[\/tex] de pl\u00e2nctons do rio.<br \/>\nExpresse o tamanho da popula\u00e7\u00e3o de peixes grandes em fun\u00e7\u00e3o da quantidade de pl\u00e2ncton, sabendo que:<br \/>\n [tex]\\qquad \\qquad f(n)=40+\\sqrt{\\dfrac{n+1}{120}}\\quad [\/tex] e [tex]\\quad g(x)=4x+5.[\/tex]<\/p>\n<p style=\"margin:10px;\">\n<p><span style=\"color:#98622F;\">&#9658;<\/span> Observe que, pelos dados do problema, [tex]n=g(x)[\/tex]; assim, basta substituir [tex]n[\/tex] por [tex]g(x)[\/tex] na express\u00e3o [tex]f(n)[\/tex], ou seja, a partir da composta [tex]f\\circ g[\/tex] teremos o tamanho da popula\u00e7\u00e3o da esp\u00e9cie de peixes grandes do rio em fun\u00e7\u00e3o da quantidade de pl\u00e2nctons. Vamos aos c\u00e1lculos:<br \/>\n[tex]\\qquad f(n)=40+\\sqrt{\\dfrac{n+1}{120}}\\\\<br \/>\n\\qquad f(g(x))=40+\\sqrt{\\dfrac{4x+5+1}{120}}\\\\<br \/>\n\\qquad f(g(x))=40+\\sqrt{\\dfrac{4x+6}{120}}\\\\<br \/>\n\\qquad \\boxed{f(g(x))=40+\\sqrt{\\dfrac{2x+3}{60}}}.[\/tex]\n<\/div>\n<p style=\"margin:20px;\">\n<div class=\"gr1 cbm5\" style=\"color: #1e1e1e; padding:20px; width:95%;font-size: 15px;line-height: 30px;\">\n<strong><span style=\"color: #8B5A2B;\">Exemplo 6:<\/span><\/strong> (<strong>PUCPR<\/strong>) Seja [tex]y=f(x)[\/tex] uma fun\u00e7\u00e3o definida no intervalo [tex][-3,6][\/tex], conforme indicado no gr\u00e1fico.<br \/>\nDeste modo, determinar o valor de [tex]f(f(2))[\/tex].<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2023\/10\/func037.png\" style=\"border:none\" alt=\"\" width=\"450\" height=\"355\" class=\"aligncenter size-full wp-image-202498\" \/><\/p>\n<p style=\"margin:10px;\">\n<p><span style=\"color:#98622F;\">&#9658;<\/span> Observando o gr\u00e1fico, temos que:<br \/>\n[tex]\\quad \\textcolor{#98622F}{(i)~}f(2)=-3[\/tex]<br \/>\n[tex]\\quad \\textcolor{#98622F}{(ii)~}f(-3)=1.[\/tex]<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2023\/10\/func037_1.png\" style=\"border:none\" alt=\"\" width=\"450\" height=\"355\" class=\"aligncenter size-full wp-image-202506\" \/><\/p>\n<p>Portanto,<br \/>\n[tex]\\quad \\boxed{f(f(2))=f(-3)=1}.[\/tex]\n<\/p><\/div>\n<p style=\"margin:20px;\">\n<div class=\"gr1 cbm5\" style=\"color: #1e1e1e; padding:20px; width:95%;font-size: 15px;line-height: 30px;\">\n<strong><span style=\"color: #8B5A2B;\">Exemplo 7:<\/span><\/strong> (<strong>UERJ<\/strong>-02) Admita os seguintes dados sobre as condi\u00e7\u00f5es ambientais de uma comunidade, com popula\u00e7\u00e3o [tex]p[\/tex], em milhares de habitantes:<br \/>\n&#8211; [tex]C[\/tex], a taxa m\u00e9dia di\u00e1ria de mon\u00f3xido de carbono no ar, em partes por milh\u00e3o, corresponde a [tex]C(p)=0,5p+1[\/tex];<br \/>\n&#8211; em um determinado tempo [tex]t[\/tex], em anos, [tex]p[\/tex] ser\u00e1 igual a [tex]p(t)=10+0,1t^2.[\/tex]<br \/>\nEm rela\u00e7\u00e3o \u00e0 taxa [tex]C:[\/tex]<br \/>\n<strong>a)<\/strong> expresse-a como uma fun\u00e7\u00e3o de tempo;<br \/>\n<strong>b)<\/strong> calcule em quantos anos essa taxa ser\u00e1 de [tex]13,2 [\/tex] partes por milh\u00e3o.<\/p>\n<p style=\"margin:10px;\">\n<p><strong>a)<\/strong> A taxa m\u00e9dia di\u00e1ria de mon\u00f3xido de carbono no ar em fun\u00e7\u00e3o da popula\u00e7\u00e3o [tex]p[\/tex] \u00e9 dada por [tex]C(p)=0,5p+1.[\/tex]<br \/>\nPor outro lado, a popula\u00e7\u00e3o [tex]p[\/tex] em um determinado tempo [tex]t[\/tex] \u00e9 dada por [tex]p(t)=10+0,1t^2.[\/tex]<br \/>\nAssim, fazendo a composi\u00e7\u00e3o das fun\u00e7\u00f5es [tex]C[\/tex] e [tex]p[\/tex], teremos a taxa [tex]C[\/tex] expressa como fun\u00e7\u00e3o do tempo [tex]t.[\/tex]<br \/>\nVamos aos c\u00e1lculos:<br \/>\n[tex]\\qquad C(t)=C(p(t))\\\\<br \/>\n\\qquad C(t)= 0,5p(t)+1\\\\<br \/>\n\\qquad C(t)= 0,5\\left(10+0,1t^2\\right)+1\\\\<br \/>\n\\qquad C(t)= \\left(5+0,05t^2\\right)+1\\\\<br \/>\n\\qquad \\boxed{C(t)= 6+0,05t^2}.[\/tex]<\/p>\n<p style=\"margin:10px;\">\n<p><strong>b)<\/strong> Fazendo [tex]C(t)=13,2 [\/tex] na express\u00e3o resultante do item anterior, segue que:<br \/>\n[tex]\\qquad C(t)=6+0,05t^2\\\\<br \/>\n\\qquad 13,2=6+0,05t^2\\\\<br \/>\n\\qquad 7,2=0,05t^2\\\\<br \/>\n\\qquad t^2=\\dfrac{7,2}{0,05}\\\\<br \/>\n\\qquad t^2=144.[\/tex]<br \/>\nComo [tex]t\\gt 0[\/tex], conclu\u00edmos que [tex]t=12[\/tex] anos.\n<\/div>\n<p style=\"margin:20px;\">\n<div class=\"gr1 cbm5\" style=\"color: #1e1e1e; padding:20px; width:95%;font-size: 15px;line-height: 30px;\">\n<strong><span style=\"color: #8B5A2B;\">Exemplo 8:<\/span><\/strong> (<strong>MACK<\/strong>-02) Na figura, temos os esbo\u00e7os dos gr\u00e1ficos das fun\u00e7\u00f5es [tex]f[\/tex] e [tex]g.[\/tex]<br \/>\nDetermine o valor da soma [tex]f(g(1))+g(f(\u20132)).[\/tex]<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2023\/10\/func038.png\"  style=\"border:none\" alt=\"\" width=\"350\" height=\"329\" class=\"aligncenter size-full wp-image-202532\" \/><\/p>\n<p style=\"margin:10px;\">\n<p><span style=\"color:#98622F;\">&#9658;<\/span> Observando o gr\u00e1fico das fun\u00e7\u00f5es [tex]f[\/tex] e [tex]g[\/tex], segue que:<br \/>\n[tex]\\quad \\textcolor{#98622F}{(i)~}g(1)=0[\/tex]<br \/>\n[tex]\\quad \\textcolor{#98622F}{(ii)~}f(-2)=0[\/tex]<br \/>\n[tex]\\quad \\textcolor{#98622F}{(iii)~}f(g(1))+g(f(\u20132))=f(0)+g(0)[\/tex]<br \/>\n[tex]\\quad \\textcolor{#98622F}{(iv)~}f(0)=0[\/tex]<br \/>\n[tex]\\quad \\textcolor{#98622F}{(v)~}g(0)=2.[\/tex]<br \/>\nAssim:<br \/>\n[tex]\\qquad f(g(1))+g(f(\u20132))=f(0)+g(0)=0+2=2.[\/tex]\n<\/div>\n<p style=\"margin:20px;\">\n<div class=\"gr1 cbm5\" style=\"color: #1e1e1e; padding:20px; width:95%;font-size: 15px;line-height: 30px;\">\n<strong><span style=\"color: #8B5A2B;\">Exemplo 9:<\/span><\/strong> (<strong>UEL<\/strong>) Se [tex]f[\/tex] e [tex]g[\/tex] s\u00e3o fun\u00e7\u00f5es de [tex]\\mathbb{R}[\/tex] em  [tex]\\mathbb{R}[\/tex] tais que [tex]f(x)=2x-1[\/tex] e [tex]f(g(x))=x^2-1[\/tex], ent\u00e3o [tex]g(x)[\/tex] \u00e9 igual a <\/p>\n<p style=\"margin:10px;\"><strong>a)<\/strong> [tex]2x^2+1\\qquad [\/tex]  <strong>b)<\/strong> [tex](x\/2)-1\\qquad [\/tex]  <strong>c)<\/strong> [tex]x^2\/2\\qquad [\/tex]  <strong>d)<\/strong> [tex]x+1\\qquad [\/tex]  <strong>e)<\/strong> [tex]x+(1\/2)[\/tex] <\/p>\n<p><span style=\"color:#98622F;\">&#9658;<\/span> Por defini\u00e7\u00e3o de fun\u00e7\u00e3o composta,<br \/>\n [tex]\\qquad f(g(x))=2g(x)-1[\/tex]<br \/>\ne pelos dados do problema,<br \/>\n [tex]\\qquad f(g(x))=x^2-1.[\/tex]<br \/>\nLogo, segue que:<br \/>\n [tex]\\qquad 2g(x)-1=x^2-1\\\\<br \/>\n\\qquad 2g(x)=x^2\\\\<br \/>\n\\qquad g(x)=\\dfrac{x^2}{2}.\\\\<br \/>\n[\/tex]<br \/>\nPortanto o item correto \u00e9 o <strong>item c<\/strong>.\n<\/div>\n<p style=\"margin:20px;\">\n<div class=\"gr1 cbm5\" style=\"color: #1e1e1e; padding:20px; width:95%;font-size: 15px;line-height: 30px;\">\n<strong><span style=\"color: #8B5A2B;\">Exemplo 10:<\/span><\/strong> (<strong>FCG<\/strong>) As fun\u00e7\u00f5es [tex]f[\/tex] e [tex]g[\/tex], de [tex]\\mathbb{R}[\/tex] em  [tex]\\mathbb{R}[\/tex], s\u00e3o definidas por [tex]f(x)=2x+3[\/tex]  e [tex]g(x)=3x+ m.[\/tex]<br \/>\n Se [tex]f(g(x))=g(f(x))[\/tex], determine [tex]f(m).[\/tex]<br \/>\n<span style=\"color:#98622F;\">&#9658;<\/span> Por defini\u00e7\u00e3o de fun\u00e7\u00e3o composta, temos que:<br \/>\n[tex]\\qquad f\\left(g(x)\\right)=2g(x)+3=2\\cdot(3x+ m)+3\\\\<br \/>\n\\qquad f\\left(g(x)\\right)=6x+2m+3[\/tex]<br \/>\ne<br \/>\n[tex]\\qquad g\\left(f(x)\\right)=3f(x)+ m=3\\cdot (2x+3)+ m\\\\<br \/>\n\\qquad g\\left(f(x)\\right)=6x+9+m.\\\\<br \/>\n[\/tex]<br \/>\nComo [tex]f(g(x))=g(f(x))[\/tex], segue que:<br \/>\n[tex] \\qquad 6x+2m+3=6x+9+m\\\\<br \/>\n\\qquad 2m+3=9+m \\\\<br \/>\n\\qquad m=6.[\/tex]<br \/>\nCom isso:<br \/>\n[tex]\\qquad \\boxed{f(6)=12+3=15}.[\/tex]\n<\/div>\n<\/div>\n<p class=\"fim\"><a title=\"Voltar.\" href=\"http:\/\/clubes.obmep.org.br\/blog\/?p=205350\"><span style=\"color: #C68445;\"><strong>Voltar para a Sala sobre Fun\u00e7\u00e3o Composta e Fun\u00e7\u00e3o Inversa<\/strong><\/span><\/a><\/p>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Clique no bot\u00e3o abaixo para visualizar a Sala. Um pouco mais de composi\u00e7\u00e3o de fun\u00e7\u00f5es Nesta Sala vamos fazer alguns exemplos mais envolvendo composi\u00e7\u00e3o de fun\u00e7\u00f5es. Exemplo 1: Considere as seguintes fun\u00e7\u00f5es: [tex]\\qquad \\begin{eqnarray*} f:\\mathbb{R} &#038;\\rightarrow &#038; \\mathbb{R}\\qquad \\qquad g:&#038; \\mathbb{R} \\rightarrow &#038; \\mathbb{R}\\\\ x&#038;\\mapsto &#038; x^2 &#038; x\\mapsto &#038; x+2 \\end{eqnarray*}[\/tex] \u00c9 poss\u00edvel definir &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/a-um-pouco-mais-de-composicao-de-funcoes\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-205356","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/205356","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=205356"}],"version-history":[{"count":3,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/205356\/revisions"}],"predecessor-version":[{"id":205359,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/205356\/revisions\/205359"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=205356"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}