\nEnunciado<\/b><\/span><\/p>\n
Considere a fun\u00e7\u00e3o $g:x\\to 1+{{\\log }_{3}}(2-5x)$.<\/p>\n
- \n
- Determine o dom\u00ednio e os zeros de $g$.<\/li>\n
- Resolva as condi\u00e7\u00f5es:
\na) $g(x)\\le 3$
\nb) $g(x)>1$<\/li>\n - Confirme, na sua calculadora, os resultados encontrados.<\/li>\n<\/ol>\nResolu\u00e7\u00e3o >><\/a><\/span><\/div><\/div>\n\n\nResolu\u00e7\u00e3o<\/b><\/span><\/p>\n
\n
Considere a fun\u00e7\u00e3o $$g:x\\to 1+{{\\log }_{3}}(2-5x)$$<\/p>\n<\/blockquote>\n
\u00ad<\/p>\n
- \n
- \u00a0Comecemos por determinar o dom\u00ednio de $g$: \\[\\begin{array}{*{35}{l}} \u00a0\u00a0 {{D}_{g}} & = & \\left\\{ x\\in \\mathbb{R}:2-5x>0 \\right\\}\u00a0 \\\\ \u00a0\u00a0 {} & = & \\left\\{ x\\in \\mathbb{R}:x<\\frac{2}{5} \\right\\}\u00a0 \\\\ \u00a0\u00a0 {} & = & \\left] -\\infty ,\\frac{2}{5} \\right[\u00a0 \\\\ \\end{array}\\] Determinemos, agora, os zeros de $g$: \\[\\begin{array}{*{35}{l}} \u00a0\u00a0 g(x)=0 & \\Leftrightarrow\u00a0 & 1+{{\\log }_{3}}(2-5x)=0\u00a0 \\\\ \u00a0\u00a0 {} & \\Leftrightarrow\u00a0 & {{\\log }_{3}}(2-5x)=-1\u00a0 \\\\ \u00a0\u00a0 {} & \\Leftrightarrow\u00a0 & \\begin{array}{*{35}{l}} \u00a0\u00a0 2-5x={{3}^{-1}} & \\wedge\u00a0 & x\\in {{D}_{g}}\u00a0 \\\\ \\end{array}\u00a0 \\\\ \u00a0\u00a0 {} & \\Leftrightarrow\u00a0 & \\begin{array}{*{35}{l}} \u00a0\u00a0 2-5x=\\frac{1}{3} & \\wedge\u00a0 & x\\in {{D}_{g}}\u00a0 \\\\ \\end{array}\u00a0 \\\\ \u00a0\u00a0 {} & \\Leftrightarrow\u00a0 & \\begin{array}{*{35}{l}} \u00a0\u00a0 6-15x=1 & \\wedge\u00a0 & x\\in {{D}_{g}}\u00a0 \\\\ \\end{array}\u00a0 \\\\ \u00a0\u00a0 {} & \\Leftrightarrow\u00a0 & \\begin{array}{*{35}{l}} \u00a0\u00a0 x=\\frac{1}{3} & \\wedge\u00a0 & x\\in \\left] -\\infty ,\\frac{2}{5} \\right[\u00a0 \\\\ \\end{array}\u00a0 \\\\ \u00a0\u00a0 {} & \\Leftrightarrow\u00a0 & x=\\frac{1}{3}\u00a0 \\\\ \\end{array}\\] A fun\u00e7\u00e3o apenas possui um zero: $x=\\frac{1}{3}$.
\n\u00ad<\/li>\n - \u00a0 a) \\[\\begin{array}{*{35}{l}} \u00a0\u00a0 g(x)\\le 3 & \\Leftrightarrow\u00a0 & 1+{{\\log }_{3}}(2-5x)\\le 3\u00a0 \\\\ \u00a0\u00a0 {} & \\Leftrightarrow\u00a0 & {{\\log }_{3}}(2-5x)\\le 2\u00a0 \\\\ \u00a0\u00a0 {} & \\Leftrightarrow\u00a0 & {{\\log }_{3}}(2-5x)\\le {{\\log }_{3}}{{3}^{2}}\u00a0 \\\\ \u00a0\u00a0 {} & \\Leftrightarrow\u00a0 & \\begin{array}{*{35}{l}} \u00a0\u00a0 2-5x\\le 9 & \\wedge\u00a0 & x\\in {{D}_{g}}\\,\\,\\,\\,(\\text{pois }x\\to {{\\log }_{3}}x\\text{\u00a0 }\\!\\!\\acute{\\mathrm{e}}\\!\\!\\text{\u00a0 estritamente crescente})\u00a0 \\\\ \\end{array}\u00a0 \\\\ \u00a0\u00a0 {} & \\Leftrightarrow\u00a0 & \\begin{array}{*{35}{l}} \u00a0\u00a0 x\\ge -\\frac{7}{5} & \\wedge\u00a0 & x\\in \\left] -\\infty ,\\frac{2}{5} \\right[\u00a0 \\\\ \\end{array}\u00a0 \\\\ \u00a0\u00a0 {} & \\Leftrightarrow\u00a0 & x\\in \\left[ -\\frac{7}{5},\\frac{2}{5} \\right[\u00a0 \\\\ \\end{array}\\] \u00a0 b) \\[\\begin{array}{*{35}{l}} \u00a0\u00a0 g(x)>1 & \\Leftrightarrow\u00a0 & 1+{{\\log }_{3}}(2-5x)>1\u00a0 \\\\ \u00a0\u00a0 {} & \\Leftrightarrow\u00a0 & {{\\log }_{3}}(2-5x)>0\u00a0 \\\\ \u00a0\u00a0 {} & \\Leftrightarrow\u00a0 & {{\\log }_{3}}(2-5x)>{{\\log }_{3}}1\u00a0 \\\\ \u00a0\u00a0 {} & \\Leftrightarrow\u00a0 & \\begin{array}{*{35}{l}} \u00a0\u00a0 2-5x>1 & \\wedge\u00a0 & x\\in {{D}_{g}}\\,\\,\\,\\,(\\text{pois }x\\to {{\\log }_{3}}x\\text{\u00a0 }\\!\\!\\acute{\\mathrm{e}}\\!\\!\\text{\u00a0 estritamente crescente})\u00a0 \\\\ \\end{array}\u00a0 \\\\ \u00a0\u00a0 {} & \\Leftrightarrow\u00a0 & \\begin{array}{*{35}{l}} \u00a0\u00a0 x<\\frac{1}{5} & \\wedge\u00a0 & x\\in \\left] -\\infty ,\\frac{2}{5} \\right[\u00a0 \\\\ \\end{array}\u00a0 \\\\ \u00a0\u00a0 {} & \\Leftrightarrow\u00a0 & x\\in \\left] -\\infty ,\\frac{1}{5} \\right[\u00a0 \\\\ \\end{array}\\]
\n\u00ad<\/li>\n - Os valores obtidos na calculadora sugerem a valida\u00e7\u00e3o dos resultados encontrados:
\n\n\n
\n ![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/01\/12p53-26G1.jpg\" "\\"G1\\"")
<\/a><\/td>\n![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/01\/12p53-26G2.jpg\" "\\"G2\\"")
<\/a><\/td>\n<\/tr>\n\n ![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/01\/12p53-26G3.jpg\" "\\"G3\\"")
<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<\/ol>\n
- \u00a0Comecemos por determinar o dom\u00ednio de $g$: \\[\\begin{array}{*{35}{l}} \u00a0\u00a0 {{D}_{g}} & = & \\left\\{ x\\in \\mathbb{R}:2-5x>0 \\right\\}\u00a0 \\\\ \u00a0\u00a0 {} & = & \\left\\{ x\\in \\mathbb{R}:x<\\frac{2}{5} \\right\\}\u00a0 \\\\ \u00a0\u00a0 {} & = & \\left] -\\infty ,\\frac{2}{5} \\right[\u00a0 \\\\ \\end{array}\\] Determinemos, agora, os zeros de $g$: \\[\\begin{array}{*{35}{l}} \u00a0\u00a0 g(x)=0 & \\Leftrightarrow\u00a0 & 1+{{\\log }_{3}}(2-5x)=0\u00a0 \\\\ \u00a0\u00a0 {} & \\Leftrightarrow\u00a0 & {{\\log }_{3}}(2-5x)=-1\u00a0 \\\\ \u00a0\u00a0 {} & \\Leftrightarrow\u00a0 & \\begin{array}{*{35}{l}} \u00a0\u00a0 2-5x={{3}^{-1}} & \\wedge\u00a0 & x\\in {{D}_{g}}\u00a0 \\\\ \\end{array}\u00a0 \\\\ \u00a0\u00a0 {} & \\Leftrightarrow\u00a0 & \\begin{array}{*{35}{l}} \u00a0\u00a0 2-5x=\\frac{1}{3} & \\wedge\u00a0 & x\\in {{D}_{g}}\u00a0 \\\\ \\end{array}\u00a0 \\\\ \u00a0\u00a0 {} & \\Leftrightarrow\u00a0 & \\begin{array}{*{35}{l}} \u00a0\u00a0 6-15x=1 & \\wedge\u00a0 & x\\in {{D}_{g}}\u00a0 \\\\ \\end{array}\u00a0 \\\\ \u00a0\u00a0 {} & \\Leftrightarrow\u00a0 & \\begin{array}{*{35}{l}} \u00a0\u00a0 x=\\frac{1}{3} & \\wedge\u00a0 & x\\in \\left] -\\infty ,\\frac{2}{5} \\right[\u00a0 \\\\ \\end{array}\u00a0 \\\\ \u00a0\u00a0 {} & \\Leftrightarrow\u00a0 & x=\\frac{1}{3}\u00a0 \\\\ \\end{array}\\] A fun\u00e7\u00e3o apenas possui um zero: $x=\\frac{1}{3}$.