Sejam<\/p>\n
\\[\\begin{matrix}
\nf:x\\to \\frac{2x+1}{{{x}^{2}}-1} & e & g:x\\to \\frac{2}{x-1}\u00a0 \\\\
\n\\end{matrix}\\]<\/p>\n
- \n
- Mostre que $f+g$ e $f-g$ s\u00e3o fun\u00e7\u00f5es racionais e determine o seu dom\u00ednio.<\/li>\n
- Resolva gr\u00e1fica e analiticamente as condi\u00e7\u00f5es:a) $f(x)\\ge 1$\n
b) $g(x)\\ge x$<\/p>\n
c) $f(x)<-\\frac{1}{2}$<\/p>\n
d) $f(x)\\ge g(x)$<\/p>\n<\/li>\n
- Determine gr\u00e1fica e analiticamente as coordenadas dos pontos do gr\u00e1fico de g que t\u00eam abcissa igual \u00e0 ordenada.<\/li>\n<\/ol>\nResolu\u00e7\u00e3o >><\/a><\/span><\/div><\/div>\n\n\nResolu\u00e7\u00e3o<\/b><\/span><\/p>\n
- \n
- Ora, ${{D}_{f}}=\\mathbb{R}\\backslash \\left\\{ -1,1 \\right\\}$ e ${{D}_{g}}=\\mathbb{R}\\backslash \\left\\{ 1 \\right\\}$.\\[f(x)+g(x)=\\frac{2x+1}{{{x}^{2}}-1}+\\frac{2}{\\underset{(x+1)}{\\mathop{x-1}}\\,}=\\frac{2x+1+2x+2}{{{x}^{2}}-1}=\\frac{4x+3}{{{x}^{2}}-1}\\]\n
\\[f(x)-g(x)=\\frac{2x+1}{{{x}^{2}}-1}-\\frac{2}{\\underset{(x+1)}{\\mathop{x-1}}\\,}=\\frac{2x+1-2x-2}{{{x}^{2}}-1}=\\frac{-1}{{{x}^{2}}-1}\\]<\/p>\n
${{D}_{f+g}}={{D}_{f-g}}=\\mathbb{R}\\backslash \\left\\{ -1,1 \\right\\}$<\/p>\n
As fun\u00e7\u00f5es $f+g$ e $f-g$ s\u00e3o fun\u00e7\u00f5es racionais, pois s\u00e3o o quociente de duas fun\u00e7\u00f5oes polinomiais, sendo o divisor diferente do polin\u00f3mio nulo:
\n\\[\\begin{matrix}
\nf+g: & \\mathbb{R}\\backslash \\left\\{ -1,1 \\right\\}\\to \\mathbb{R}\u00a0 \\\\
\n{} & x\\to \\frac{4x+3}{{{x}^{2}}-1}\u00a0 \\\\
\n\\end{matrix}\\]<\/p>\n\\[\\begin{matrix}
\nf-g: & \\mathbb{R}\\backslash \\left\\{ -1,1 \\right\\}\\to \\mathbb{R}\u00a0 \\\\
\n{} & x\\to \\frac{-1}{{{x}^{2}}-1}\u00a0 \\\\
\n\\end{matrix}\\]<\/p>\n<\/li>\n - a) $f(x)\\ge 1$
\n\\[\\begin{array}{*{35}{l}}
\nf(x)\\ge 1 & \\Leftrightarrow\u00a0 & \\frac{2x+1}{{{x}^{2}}-1}\\ge 1\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & \\frac{2x+1-{{x}^{2}}+1}{{{x}^{2}}-1}\\ge 0\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & \\frac{-{{x}^{2}}+2x+2}{{{x}^{2}}-1}\\ge 0\u00a0 \\\\
\n\\end{array}\\]<\/p>\nC\u00e1lculos auxiliares:
\n\\[\\begin{array}{*{35}{l}}
\n-{{x}^{2}}+2x+2=0 & \\Leftrightarrow\u00a0 & x=\\frac{-2\\pm \\sqrt{4+8}}{-2}\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & x=\\frac{-2\\pm 2\\sqrt{3}}{-2}\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & x=1-\\sqrt{3}\\vee x=1+\\sqrt{3}\u00a0 \\\\
\n\\end{array}\\]<\/p>\n
- Ora, ${{D}_{f}}=\\mathbb{R}\\backslash \\left\\{ -1,1 \\right\\}$ e ${{D}_{g}}=\\mathbb{R}\\backslash \\left\\{ 1 \\right\\}$.\\[f(x)+g(x)=\\frac{2x+1}{{{x}^{2}}-1}+\\frac{2}{\\underset{(x+1)}{\\mathop{x-1}}\\,}=\\frac{2x+1+2x+2}{{{x}^{2}}-1}=\\frac{4x+3}{{{x}^{2}}-1}\\]\n
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