\nEnunciado<\/b><\/span><\/p>\n
Considere a fun\u00e7\u00e3o real de vari\u00e1vel real assim definida: $$f(x) = 1 + 2\\cos \\left( {x – \\frac{\\pi }{3}} \\right)$$<\/p>\n
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- O gr\u00e1fico seguinte representa a fun\u00e7\u00e3o cosseno. Explique como a partir dele obt\u00e9m o gr\u00e1fico de $f$.
![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12pag125-2.jpg\" "\\"Gr\u00e1fico\\"")
<\/a><\/li>\n- Calcule o valor exato de $f\\left( {\\frac{{7\\pi }}{2}} \\right) – f\\left( {\\frac{{7\\pi }}{6}} \\right)$.<\/li>\n
- Determine o contradom\u00ednio da fun\u00e7\u00e3o dada.<\/li>\n
- Determine uma express\u00e3o geral dos zeros da fun\u00e7\u00e3o.<\/li>\n
- Averigue se $f(x + 2k\\pi ) = f(x),\\forall x \\in \\mathbb{R}$, com $k \\in \\mathbb{Z}$. O que pode concluir?<\/li>\n<\/ol>\n
Resolu\u00e7\u00e3o >><\/a><\/span><\/div><\/div>\n\n\nResolu\u00e7\u00e3o<\/b><\/span><\/p>\n- \n
- O gr\u00e1fico da fun\u00e7\u00e3o $f$ pode ser obtido a partir do gr\u00e1fico dado atrav\u00e9s de uma transla\u00e7\u00e3o associada ao vetor $\\overrightarrow u\u00a0 = \\left( {\\frac{\\pi }{3},0} \\right)$, seguida de alongamento vertical de fator 2 e, por fim, de uma transla\u00e7\u00e3o associada ao vetor $\\overrightarrow v\u00a0 = \\left( {0,1} \\right)$.
![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12pag125-2.png\" "\\"Gr\u00e1fico\\"")
<\/a><\/li>\n- Ora, $$\\begin{array}{*{20}{l}} \u00a0 {f\\left( {\\frac{{7\\pi }}{2}} \\right) – f\\left( {\\frac{{7\\pi }}{6}} \\right)}& = &{1 + 2\\cos \\left( {\\frac{{7\\pi }}{2} – \\frac{\\pi }{3}} \\right) – \\left[ {1 + 2\\cos \\left( {\\frac{{7\\pi }}{6} – \\frac{\\pi }{3}} \\right)} \\right]} \\\\ \u00a0 {}& = &{2\\cos \\left( {\\frac{{19\\pi }}{6}} \\right) – 2\\cos \\left( {\\frac{{5\\pi }}{6}} \\right)} \\\\ \u00a0 {}& = &{2\\cos \\left( {2\\pi\u00a0 + \\pi\u00a0 + \\frac{\\pi }{6}} \\right) – 2\\cos \\left( {\\pi\u00a0 – \\frac{\\pi }{6}} \\right)} \\\\ \u00a0 {}& = &{ – 2\\cos \\left( {\\frac{\\pi }{6}} \\right) + 2\\cos \\left( { – \\frac{\\pi }{6}} \\right)} \\\\ \u00a0 {}& = &{ – 2\\cos \\left( {\\frac{\\pi }{6}} \\right) + 2\\cos \\left( {\\frac{\\pi }{6}} \\right)} \\\\ \u00a0 {}& = &0 \\end{array}$$
\n\u00ad<\/li>\n- O contradom\u00ednio da fun\u00e7\u00e3o $x \\to \\cos \\left( {x – \\frac{\\pi }{3}} \\right)$, de dom\u00ednio $\\mathbb{R}$, \u00e9 $\\left[ { – 1,1} \\right]$. Assim, o contradom\u00ednio da fun\u00e7\u00e3o $x \\to 2\\cos \\left( {x – \\frac{\\pi }{3}} \\right)$ \u00e9 $\\left[ { – 2,2} \\right]$ e, consequentemente, ser\u00e1 $D{‘_f} = \\left[ { – 1,3} \\right]$.
\n\u00ad<\/li>\n- Ora, $$\\begin{array}{*{20}{l}} \u00a0 {f(x) = 0}& \\Leftrightarrow &{1 + 2\\cos \\left( {x – \\frac{\\pi }{3}} \\right) = 0} \\\\ \u00a0 {}& \\Leftrightarrow &{\\cos \\left( {x – \\frac{\\pi }{3}} \\right) =\u00a0 – \\frac{1}{2}} \\\\ \u00a0 {}& \\Leftrightarrow &{x – \\frac{\\pi }{3} =\u00a0 \\pm \\frac{{2\\pi }}{3} + 2k\\pi ,k \\in \\mathbb{Z}} \\\\ \u00a0 {}& \\Leftrightarrow &{\\begin{array}{*{20}{l}} \u00a0 {x =\u00a0 – \\frac{\\pi }{3} + 2k\\pi }& \\vee &{x = \\pi\u00a0 + 2k\\pi ,k \\in \\mathbb{Z}} \\end{array}} \\end{array}$$
\n\u00ad<\/li>\n- Como $$\\begin{array}{*{20}{l}} \u00a0 {f(x + 2k\\pi )}& = &{1 + 2\\cos \\left( {x + 2k\\pi\u00a0 – \\frac{\\pi }{3}} \\right)} \\\\ \u00a0 {}& = &{1 + 2\\cos \\left( {x – \\frac{\\pi }{3}} \\right)} \\\\ \u00a0 {}& = &{f(x)} \\end{array}$$ ent\u00e3o $f(x + 2k\\pi ) = f(x),\\forall x \\in \\mathbb{R}$, com $k \\in \\mathbb{Z}$, o que permite concluir que $f$ \u00e9 uma fun\u00e7\u00e3o peri\u00f3dica com per\u00edodo positivo m\u00ednimo $2\\pi $.<\/li>\n<\/ol>\n
- Ora, $$\\begin{array}{*{20}{l}} \u00a0 {f\\left( {\\frac{{7\\pi }}{2}} \\right) – f\\left( {\\frac{{7\\pi }}{6}} \\right)}& = &{1 + 2\\cos \\left( {\\frac{{7\\pi }}{2} – \\frac{\\pi }{3}} \\right) – \\left[ {1 + 2\\cos \\left( {\\frac{{7\\pi }}{6} – \\frac{\\pi }{3}} \\right)} \\right]} \\\\ \u00a0 {}& = &{2\\cos \\left( {\\frac{{19\\pi }}{6}} \\right) – 2\\cos \\left( {\\frac{{5\\pi }}{6}} \\right)} \\\\ \u00a0 {}& = &{2\\cos \\left( {2\\pi\u00a0 + \\pi\u00a0 + \\frac{\\pi }{6}} \\right) – 2\\cos \\left( {\\pi\u00a0 – \\frac{\\pi }{6}} \\right)} \\\\ \u00a0 {}& = &{ – 2\\cos \\left( {\\frac{\\pi }{6}} \\right) + 2\\cos \\left( { – \\frac{\\pi }{6}} \\right)} \\\\ \u00a0 {}& = &{ – 2\\cos \\left( {\\frac{\\pi }{6}} \\right) + 2\\cos \\left( {\\frac{\\pi }{6}} \\right)} \\\\ \u00a0 {}& = &0 \\end{array}$$