\n| $f(x)$<\/td>\n | M\u00e1x. rel.<\/td>\n | $ \\searrow $<\/td>\n | M\u00edn abs.<\/td>\n | $ \\nearrow $<\/td>\n | M\u00e1x. abs.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n \u00c9 de 10 metros a dist\u00e2ncia ao primeiro poste do ponto do fio mais pr\u00f3ximo do solo.
\n\u00ad<\/li>\n Ora, $$\\begin{array}{*{20}{l}}
\n{f(x) = 15}& \\Leftrightarrow &{5\\left( {{e^{1 – 0,1x}} + {e^{0,1x – 1}}} \\right) = 15} \\\\
\n{}& \\Leftrightarrow &{{e^{1 – 0,1x}} + {e^{0,1x – 1}} = 3} \\\\
\n{}& \\Leftrightarrow &{{e^{1 – 0,1x}} + {e^{0,1x – 1}} – 3 = 0}
\n\\end{array}$$
\nFazendo $1 – 0,1x = y$, temos:
\n$$\\begin{array}{*{20}{l}}
\n{{e^y} + {e^{ – y}} – 3 = 0}& \\Leftrightarrow &{{e^y} \\times {e^y} + {e^y} \\times {e^{ – y}} – 3 \\times {e^y} = 0} \\\\
\n{}& \\Leftrightarrow &{{{\\left( {{e^y}} \\right)}^2} – 3 \\times {e^y} + 1 = 0} \\\\
\n{}& \\Leftrightarrow &{{e^y} = \\frac{{3 \\mp \\sqrt {9 – 4} }}{2}} \\\\
\n{}& \\Leftrightarrow &{\\begin{array}{*{20}{l}}
\n{{e^y} = \\frac{{3 – \\sqrt 5 }}{2}}& \\vee &{{e^y} = \\frac{{3 + \\sqrt 5 }}{2}}
\n\\end{array}} \\\\
\n{}& \\Leftrightarrow &{\\begin{array}{*{20}{l}}
\n{y = \\ln \\frac{{3 – \\sqrt 5 }}{2}}& \\vee &{y = \\ln \\frac{{3 + \\sqrt 5 }}{2}}
\n\\end{array}}
\n\\end{array}$$
\nDado que $1 – 0,1x = y \\Leftrightarrow x = 10 + 10y$, vem: $$\\begin{array}{*{20}{l}}
\n{f(x) = 15}& \\Leftrightarrow &{\\begin{array}{*{20}{l}}
\n{x = 10 + 10\\ln \\frac{{3 – \\sqrt 5 }}{2}}& \\vee &{x = 10 + 10\\ln \\frac{{3 + \\sqrt 5 }}{2}}
\n\\end{array}}
\n\\end{array}$$
\nA dist\u00e2ncia ao primeiro poste dos pontos do fio que se encontrem a 15 metros do solo \u00e9 ${x_1} = 10 + 10\\ln \\frac{{3 – \\sqrt 5 }}{2} \\approx 0,4$ e ${x_2} = 10 + 10\\ln \\frac{{3 + \\sqrt 5 }}{2} \\approx 19,6$ metros, respetivamente.
\n\u00ad<\/li>\n<\/ol>\n |
![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12pag230-96.jpg\" "\\"Fio")
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