Para que um rem\u00e9dio produza o efeito desejado, a sua concentra\u00e7\u00e3o na corrente sangu\u00ednea deve estar acima de um certo valor, o n\u00edvel terap\u00eautico m\u00ednimo.<\/p>\n
Suponhamos que a concentra\u00e7\u00e3o c<\/strong> de um rem\u00e9dio, t<\/strong> horas ap\u00f3s ser ingerido, \u00e9 dada, em mg\/l, por: \\[c(t)=\\frac{20t}{{{t}^{2}}+4}\\]<\/p>\n Se o n\u00edvel terap\u00eautico m\u00ednimo \u00e9 de 4 mg\/l, determine quando este n\u00edvel \u00e9 excedido.<\/p>\n Come\u00e7ando por uma resolu\u00e7\u00e3o gr\u00e1fica, temos:<\/p>\n O n\u00edvel terap\u00eautico m\u00ednimo de 4 mg\/l \u00e9 ultrapassado entre a 1.\u00aa e 4.\u00aa horas ap\u00f3s o rem\u00e9dio ter sido ingerido.<\/p>\n Resolvendo analiticamente, vem:<\/p>\n \\[\\begin{array}{*{35}{l}} C\u00e1lculos auxiliares:<\/p>\n \\[\\begin{array}{*{35}{l}}![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/02\/11-pag187-16-1.jpg\" "\\"J1\\"")
<\/a>\u00a0 ![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/02\/11-pag187-16-2.jpg\" "\\"G1\\"")
<\/a>\u00a0 ![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/02\/11-pag187-16-3.jpg\" "\\"G2\\"")
<\/a><\/p>\n
\nc(t)>4 & \\Leftrightarrow\u00a0 & \\begin{array}{*{35}{l}}
\n\\frac{20t}{{{t}^{2}}+4}>4 & \\wedge\u00a0 & t\\ge 0\u00a0 \\\\
\n\\end{array}\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & \\begin{array}{*{35}{l}}
\n\\frac{20t-4{{t}^{2}}-16}{{{t}^{2}}+4}>0 & \\wedge\u00a0 & t\\ge 0\u00a0 \\\\
\n\\end{array}\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & \\begin{array}{*{35}{l}}
\n20t-4{{t}^{2}}-16>0 & \\wedge\u00a0 & t\\ge 0\\,(pois\\,{{t}^{2}}+4>0)\u00a0 \\\\
\n\\end{array}\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & \\begin{array}{*{35}{l}}
\n{{t}^{2}}-5t+4<0 & \\wedge\u00a0 & t\\ge 0\u00a0 \\\\
\n\\end{array}\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & \\begin{array}{*{35}{l}}
\n(t-1)(t-4)<0 & \\wedge\u00a0 & t\\ge 0\u00a0 \\\\
\n\\end{array}\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & t\\in \\left] 1,4 \\right[\u00a0 \\\\
\n\\end{array}\\]<\/p>\n
\n{{t}^{2}}-5t+4=0 & \\Leftrightarrow\u00a0 & t=\\frac{5\\pm \\sqrt{25-16}}{2}\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & t=\\frac{5\\pm 3}{2}\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & t=1\\vee t=4\u00a0 \\\\
\n\\end{array}\\]<\/p>\n