Defina, sem usar o s\u00edmbolo de m\u00f3dulo, e represente graficamente, cada uma das seguintes fun\u00e7\u00f5es:<\/p>\n
- \n
- $f(x) = \\left| {x – 1} \\right| + 2$<\/li>\n
- $g(x) =\u00a0 – \\left| {3{x^2} – 2x – 1} \\right|$<\/li>\n
- $h(x) =\u00a0 – \\left| {x\\left( {x – 2} \\right)\\left( {x + 1} \\right)} \\right|$<\/li>\n<\/ol>\nResolu\u00e7\u00e3o >><\/a><\/span><\/div><\/div>\n\n\nResolu\u00e7\u00e3o<\/b><\/span><\/p>\n
- \n
- \n
![\"Gr\u00e1fico](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/02\/graf1.png\")
<\/a>Gr\u00e1fico de $f$<\/p><\/div>\n
Tem-se sucessivamente:
\n\\[\\begin{array}{*{20}{l}}
\n{f(x)}& = &{\\left| {x – 1} \\right| + 2} \\\\
\n{}& = &{\\left\\{ {\\begin{array}{*{20}{l}}
\n{\\left( {x – 1} \\right) + 2}& \\Leftarrow &{x – 1 \\geqslant 0} \\\\
\n{ – \\left( {x – 1} \\right) + 2}& \\Leftarrow &{x – 1 < 0}
\n\\end{array}} \\right.} \\\\
\n{}& = &{\\left\\{ {\\begin{array}{*{20}{l}}
\n{ – x + 3}& \\Leftarrow &{x < 1} \\\\
\n{x + 1}& \\Leftarrow &{x \\geqslant 1}
\n\\end{array}} \\right.}
\n\\end{array}\\]<\/p>\nPara a representa\u00e7\u00e3o gr\u00e1fica de $f$ recorreu-se aos gr\u00e1ficos das fun\u00e7\u00f5es auxiliares ${y_1} =\u00a0 – x + 3$ e ${y_2} = x + 1$.<\/p>\n<\/li>\n
- Comecemos por calcular os zeros da fun\u00e7\u00e3o quadr\u00e1tica $y = 3{x^2} – 2x – 1$:
\n\\[\\begin{array}{*{20}{l}}
\n{3{x^2} – 2x – 1 = 0}& \\Leftrightarrow &{x = \\frac{{2 \\mp \\sqrt {4 + 12} }}{6}} \\\\
\n{}& \\Leftrightarrow &{\\begin{array}{*{20}{l}}
\n{x =\u00a0 – \\frac{1}{3}}& \\vee &{x = 1}
\n\\end{array}}
\n\\end{array}\\]<\/p>\n![\"Gr\u00e1fico](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/02\/graf2.png\")
<\/a>Gr\u00e1fico de $g$<\/p><\/div>\n
Tendo em considera\u00e7\u00e3o que o gr\u00e1fico de $y = 3{x^2} – 2x – 1$ \u00e9 uma par\u00e1bola com a concavidade voltada para cima e que interseta o eixo $Ox$ nos pontos de abcissas ${ – \\frac{1}{3}}$ e $1$, vem:
\n\\[\\begin{array}{*{20}{l}}
\n{g(x)}& = &{ – \\left| {3{x^2} – 2x – 1} \\right|} \\\\
\n{}& = &{\\left\\{ {\\begin{array}{*{20}{l}}
\n{ – \\left( { – \\left( {3{x^2} – 2x – 1} \\right)} \\right)}& \\Leftarrow &{x \\in \\left] { – \\frac{1}{3},1} \\right[} \\\\
\n{ – \\left( {3{x^2} – 2x – 1} \\right)}& \\Leftarrow &{x \\in \\left] { – \\infty , – \\frac{1}{3}} \\right] \\cup \\left[ {1, + \\infty } \\right[}
\n\\end{array}} \\right.} \\\\
\n{}& = &{\\left\\{ {\\begin{array}{*{20}{l}}
\n{3{x^2} – 2x – 1}& \\Leftarrow &{x \\in \\left] { – \\frac{1}{3},1} \\right[} \\\\
\n{ – 3{x^2} + 2x + 1}& \\Leftarrow &{x \\in \\left] { – \\infty , – \\frac{1}{3}} \\right] \\cup \\left[ {1, + \\infty } \\right[}
\n\\end{array}} \\right.}
\n\\end{array}\\]<\/p>\nPara a representa\u00e7\u00e3o gr\u00e1fica de $g$ recorreu-se aos gr\u00e1ficos das fun\u00e7\u00f5es auxiliares ${y_1} = 3{x^2} – 2x – 1$ e ${y_2} =\u00a0 – 3{x^2} + 2x + 1$.<\/p>\n<\/li>\n
- \n
Comecemos por estudar o sinal da fun\u00e7\u00e3o $y = x\\left( {x – 2} \\right)\\left( {x + 1} \\right)$:
\n\\[\\begin{array}{*{20}{c}}
\nx&{ – \\infty }&{}&{}&{ – 1}&{}&0&{}&2&{}&{}&{ + \\infty } \\\\
\n\\hline
\nx&{}& – &{}& – & – &0& + & + &{}& + &{} \\\\
\n\\hline
\n{x – 2}&{}& – &{}& – & – & – & – &0&{}& + &{} \\\\
\n\\hline
\n{x + 1}&{}& – &{}&0& + & + & + & + &{}& + &{} \\\\
\n\\hline
\ny&{}& – &{}&0& + &0& – &0&{}& + &{}
\n\\end{array}\\]<\/p>\n - Comecemos por calcular os zeros da fun\u00e7\u00e3o quadr\u00e1tica $y = 3{x^2} – 2x – 1$:
- \n
![\"Gr\u00e1fico](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2014\/02\/graf3.png\")
<\/a>