{"id":11529,"date":"2013-02-19T21:16:03","date_gmt":"2013-02-19T21:16:03","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=11529"},"modified":"2022-01-14T17:36:26","modified_gmt":"2022-01-14T17:36:26","slug":"tres-funcoes","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=11529","title":{"rendered":"Tr\u00eas fun\u00e7\u00f5es"},"content":{"rendered":"<p><ul id='GTTabs_ul_11529' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_11529' class='GTTabs_curr'><a  id=\"11529_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_11529' ><a  id=\"11529_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_11529'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Dadas as fun\u00e7\u00f5es<\/p>\n<p>$$\\begin{array}{*{20}{c}}<br \/>\n{\\begin{array}{*{20}{l}}<br \/>\n{f:}&amp;{\\mathbb{R} \\to \\mathbb{R}} \\\\<br \/>\n{}&amp;{x \\to 2x + 5}<br \/>\n\\end{array}}&amp;{}&amp;{\\begin{array}{*{20}{l}}<br \/>\n{g:}&amp;{\\mathbb{R} \\to \\mathbb{R}} \\\\<br \/>\n{}&amp;{x \\to \\frac{2}{5}x + \\frac{1}{5}}<br \/>\n\\end{array}}&amp;{}&amp;{\\begin{array}{*{20}{l}}<br \/>\n{h:}&amp;{\\mathbb{R} \\to \\mathbb{R}} \\\\<br \/>\n{}&amp;{x \\to 4{x^2} &#8211; 36x}<br \/>\n\\end{array}}<br \/>\n\\end{array}$$<\/p>\n<\/p>\n<ol>\n<li>Determine a imagem de $0$, $ &#8211; 1$ e $\\frac{3}{2}$ pela fun\u00e7\u00e3o $f$.<\/li>\n<li>Qual(ais) o(s) objeto(s) que tem(t\u00eam) imagem $3$ pela fun\u00e7\u00e3o $f$.<\/li>\n<li>Represente graficamente as fun\u00e7\u00f5es dadas.<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_11529' onClick='GTTabs_show(1,11529)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_11529'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<blockquote>\n<p>$$\\begin{array}{*{20}{c}}<br \/>\n{\\begin{array}{*{20}{l}}<br \/>\n{f:}&amp;{\\mathbb{R} \\to \\mathbb{R}} \\\\<br \/>\n{}&amp;{x \\to 2x + 5}<br \/>\n\\end{array}}&amp;{}&amp;{\\begin{array}{*{20}{l}}<br \/>\n{g:}&amp;{\\mathbb{R} \\to \\mathbb{R}} \\\\<br \/>\n{}&amp;{x \\to \\frac{2}{5}x + \\frac{1}{5}}<br \/>\n\\end{array}}&amp;{}&amp;{\\begin{array}{*{20}{l}}<br \/>\n{h:}&amp;{\\mathbb{R} \\to \\mathbb{R}} \\\\<br \/>\n{}&amp;{x \\to 4{x^2} &#8211; 36x}<br \/>\n\\end{array}}<br \/>\n\\end{array}$$<\/p>\n<\/blockquote>\n<ol>\n<li>\u00a0A imagem de $0$ por $f$ \u00e9 $f\\left( 0 \\right) = 2 \\times 0 + 5 = 5$.\n<p>A imagem de $0$ por $f$ \u00e9 $f\\left( { &#8211; 1} \\right) = 2 \\times \\left( { &#8211; 1} \\right) + 5 = 3$.<\/p>\n<p>A imagem de $0$ por $f$ \u00e9 $f\\left( {\\frac{3}{2}} \\right) = 2 \\times \\frac{3}{2} + 5 = 8$.<\/p>\n<\/li>\n<li>Como $$\\begin{array}{*{20}{l}}<br \/>\n{g(x) = 3}&amp; \\Leftrightarrow &amp;{\\frac{2}{5}x + \\frac{1}{5} = 3} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{2x + 1 = 15} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{x = 7}<br \/>\n\\end{array}$$ apenas o objeto $7$ tem imagem $3$ pela fun\u00e7\u00e3o $g$.<\/p>\n<\/li>\n<li>Os gr\u00e1ficos das fun\u00e7\u00f5es $f$ e $g$ s\u00e3o retas obl\u00edquas, pois estas s\u00e3o definidas por uma equa\u00e7\u00e3o da forma $y = mx + b$, com $m \\ne 0$.<br \/>\nPor isso, para construir os gr\u00e1ficos destas fun\u00e7\u00f5es bastar\u00e1 determinar dois pontos de cada uma dessas retas:<\/p>\n<table border=\"0\" align=\"center\">\n<tbody>\n<tr>\n<td>$f\\left( 0 \\right) = 2 \\times 0 + 5 = 5$ e $f\\left( { &#8211; 1} \\right) = 2 \\times \\left( { &#8211; 1} \\right) + 5 = 3$, logo $A\\left( {0,5} \\right)$ e $B\\left( { &#8211; 1,3} \\right)$ s\u00e3o dois pontos do gr\u00e1fico de $f$.<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<\/tr>\n<tr>\n<td>$g\\left( 2 \\right) = \\frac{2}{5} \\times 2 + \\frac{1}{5} = 1$ e $g\\left( { &#8211; 3} \\right) = \\frac{2}{5} \\times \\left( { &#8211; 3} \\right) + \\frac{1}{5} = &#8211; 1$, logo $C\\left( {2,1} \\right)$ e $D\\left( { &#8211; 3, &#8211; 1} \\right)$ s\u00e3o dois pontos do gr\u00e1fico de $g$.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Como se sabe, o gr\u00e1fico da fun\u00e7\u00e3o $h$ \u00e9 uma par\u00e1bola. Podemos come\u00e7ar por determinar alguns pontos do seu gr\u00e1fico:<\/p>\n<table border=\"0\" align=\"center\">\n<tbody>\n<tr>\n<td>Como $$\\begin{array}{*{20}{l}}<br \/>\n{h\\left( x \\right) = 0}&amp; \\Leftrightarrow &amp;{4{x^2} &#8211; 36x = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{4x\\left( {x &#8211; 9} \\right) = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}<br \/>\n{4x = 0}&amp; \\vee &amp;{x &#8211; 9 = 0}<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{c}}<br \/>\n{x = 0}&amp; \\vee &amp;{x = 9}<br \/>\n\\end{array}}<br \/>\n\\end{array}$$ ent\u00e3o $O\\left( {0,0} \\right)$ e $E\\left( {9,0} \\right)$ s\u00e3o dois pontos do gr\u00e1fico de $h$.<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<\/tr>\n<tr>\n<td>$h\\left( { &#8211; 2} \\right) = 4 \\times {\\left( { &#8211; 2} \\right)^2} &#8211; 36 \\times \\left( { &#8211; 2} \\right) = 88$ e $h\\left( {11} \\right) = 4 \\times {11^2} &#8211; 36 \\times 11 = 88$, logo\u00a0$F\\left( { &#8211; 2,88} \\right)$ e $G\\left( {11,88} \\right)$ s\u00e3o dois pontos do gr\u00e1fico de $h$.<\/td>\n<\/tr>\n<tr>\n<td>$h\\left( { &#8211; 1} \\right) = 4 \\times {\\left( { &#8211; 1} \\right)^2} &#8211; 36 \\times \\left( { &#8211; 1} \\right) = 40$ e $h\\left( {10} \\right) = 4 \\times {10^2} &#8211; 36 \\times 10 = 40$, logo $H\\left( { &#8211; 1,40} \\right)$ e $I\\left( {10,40} \\right)$ s\u00e3o dois pontos do gr\u00e1fico de $h$.<\/td>\n<\/tr>\n<tr>\n<td>$h\\left( 2 \\right) = 4 \\times {2^2} &#8211; 36 \\times 2 =\u00a0 &#8211; 56$ e $h\\left( 7 \\right) = 4 \\times {7^2} &#8211; 36 \\times 7 =\u00a0 &#8211; 56$, logo $J\\left( {2, &#8211; 56} \\right)$ e $L\\left( {7, &#8211; 56} \\right)$ s\u00e3o dois pontos do gr\u00e1fico de $h$.<\/td>\n<\/tr>\n<tr>\n<td>$h\\left( 4 \\right) = 4 \\times {4^2} &#8211; 36 \\times 4 =\u00a0 &#8211; 80$ e $h\\left( 5 \\right) = 4 \\times {5^2} &#8211; 36 \\times 5 =\u00a0 &#8211; 80$, logo $M\\left( {4, &#8211; 80} \\right)$ e $N\\left( {5, &#8211; 80} \\right)$ s\u00e3o dois pontos do gr\u00e1fico de $h$.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<\/ol>\n<p style=\"text-align: center;\"><script src=\"https:\/\/cdn.geogebra.org\/apps\/deployggb.js\"><\/script>\r\n<div id=\"ggbApplet\" style=\"margin: 0 auto;\"><\/div>\r\n<script>\r\nvar parameters = {\r\n\"id\": \"ggbApplet\",\r\n\"width\":727,\r\n\"height\":506,\r\n\"showMenuBar\":false,\r\n\"showAlgebraInput\":false,\r\n\"showToolBar\":false,\r\n\"customToolBar\":\"0 39 59 || 1 501 67 , 5 19 , 72 | 2 15 45 , 18 65 , 7 37 | 4 3 8 9 , 13 44 , 58 , 47 || 16 51 64 , 70 | 10 34 53 11 , 24  20 22 , 21 23 | 55 56 57 , 12 || 36 46 , 38 49 50 , 71 | 30 29 54 32 31 33 | 17 26 62 , 14 66 68 | 25 52 60 61 || 40 41 42 , 27 28 35 , 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