Caraterize a fun\u00e7\u00e3o inversa de cada uma das fun\u00e7\u00f5es definidas por:<\/p>\n
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- $f:x\\to 1+{{2}^{x}}$<\/li>\n
- $g:x\\to {{\\log }_{2}}(3-5x)$<\/li>\n
- $h:x\\to 4-3{{e}^{-x+2}}$<\/li>\n
- $j:x\\to 4-\\ln (1-2x)$<\/li>\n<\/ul>\nR1 >><\/a><\/span><\/div><\/div>\n\n\nR1<\/b><\/span><\/p>\n
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$f:x\\to 1+{{2}^{x}}$<\/p><\/blockquote>\n<\/li>\n<\/ul>\n
Comecemos por determinar o dom\u00ednio da fun\u00e7\u00e3o:<\/p>\n
$$\\begin{array}{*{35}{l}}
\n{{D}_{f}}=D{{‘}_{{{f}^{-1}}}} & = & \\left\\{ x\\in \\mathbb{R}:x\\in \\mathbb{R} \\right\\}\u00a0 \\\\
\n{} & = & \\mathbb{R}\u00a0 \\\\
\n\\end{array}$$<\/p>\n\u00a0Ora, $$\\begin{array}{*{35}{l}}
\ny=1+{{2}^{x}} & \\Leftrightarrow\u00a0 & {{2}^{x}}=y-1\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & x={{\\log }_{2}}(y-1)\u00a0 \\\\
\n\\end{array}$$<\/p>\nLogo, o dom\u00ednio da fun\u00e7\u00e3o inversa \u00e9: \\[\\begin{array}{*{35}{l}}
\n{{D}_{{{f}^{-1}}}}=D{{‘}_{f}} & = & \\left\\{ x\\in \\mathbb{R}:x-1>0 \\right\\}\u00a0 \\\\
\n{} & = & \\left] 1,+\\infty\u00a0 \\right[\u00a0 \\\\
\n\\end{array}\\]<\/p>\nPortanto, a fun\u00e7\u00e3o inversa \u00e9 caraterizada por: \\[\\begin{array}{*{35}{l}}
\n{{f}^{-1}}: & \\left] 1,+\\infty\u00a0 \\right[\\to \\mathbb{R}\u00a0 \\\\
\n{} & x\\to {{\\log }_{2}}(x-1)\u00a0 \\\\
\n\\end{array}\\]![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/01\/12p55-28a.jpg\" "\\"Fun\u00e7\u00e3o")
<\/a><\/p>\n<< Enunciado<\/a><\/span>R2 >><\/a><\/span><\/div><\/div>\n\n\nR2<\/b><\/span><\/p>\n- \n
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$g:x\\to {{\\log }_{2}}(3-5x)$<\/p><\/blockquote>\n<\/li>\n<\/ul>\n
Comecemos por determinar o dom\u00ednio da fun\u00e7\u00e3o:<\/p>\n
\\[\\begin{array}{*{35}{l}}
\n{{D}_{g}}=D{{‘}_{{{g}^{-1}}}} & = & \\left\\{ x\\in \\mathbb{R}:3-5x>0 \\right\\}\u00a0 \\\\
\n{} & = & \\left\\{ x\\in \\mathbb{R}:x<\\frac{3}{5} \\right\\}\u00a0 \\\\
\n{} & = & \\left] -\\infty ,\\frac{3}{5} \\right[\u00a0 \\\\
\n\\end{array}\\]<\/p>\n\u00a0Ora, \\[\\begin{array}{*{35}{l}}
\ny={{\\log }_{2}}(3-5x) & \\Leftrightarrow\u00a0 & 3-5x={{2}^{y}}\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & x=\\frac{3-{{2}^{y}}}{5}\u00a0 \\\\
\n\\end{array}\\]<\/p>\nLogo, o dom\u00ednio da fun\u00e7\u00e3o inversa \u00e9: \\[\\begin{array}{*{35}{l}}
\n{{D}_{{{g}^{-1}}}}=D{{‘}_{g}} & = & \\left\\{ x\\in \\mathbb{R}:x\\in \\mathbb{R} \\right\\}\u00a0 \\\\
\n{} & = & \\mathbb{R}\u00a0 \\\\
\n\\end{array}\\]<\/p>\nPortanto, a fun\u00e7\u00e3o inversa \u00e9 caraterizada por: \\[\\begin{array}{*{35}{l}}
\n{{g}^{-1}}: & \\mathbb{R}\\to \\left] -\\infty ,\\frac{3}{5} \\right[\u00a0 \\\\
\n{} & x\\to \\frac{3-{{2}^{x}}}{5}\u00a0 \\\\
\n\\end{array}\\]![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/01\/12p55-28b.jpg\" "\\"Fun\u00e7\u00e3o")
<\/a><\/p>\n<< R1<\/a><\/span>R3 >><\/a><\/span><\/div><\/div>\n\n\nR3<\/b><\/span><\/p>\n- \n
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$h:x\\to 4-3{{e}^{-x+2}}$<\/p><\/blockquote>\n<\/li>\n<\/ul>\n
Comecemos por determinar o dom\u00ednio da fun\u00e7\u00e3o:<\/p>\n
\\[\\begin{array}{*{35}{l}}
\n{{D}_{h}}=D{{‘}_{{{h}^{-1}}}} & = & \\left\\{ x\\in \\mathbb{R}:(-x+2)\\in \\mathbb{R} \\right\\}\u00a0 \\\\
\n{} & = & \\mathbb{R}\u00a0 \\\\
\n\\end{array}\\]<\/p>\n\u00a0Ora, \\[\\begin{array}{*{35}{l}}
\ny=4-3{{e}^{-x+2}} & \\Leftrightarrow\u00a0 & {{e}^{-x+2}}=\\frac{4-y}{3}\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & -x+2=\\ln \\left( \\frac{4-y}{3} \\right)\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & x=2-\\ln \\left( \\frac{4-y}{3} \\right)\u00a0 \\\\
\n\\end{array}\\]<\/p>\nLogo, o dom\u00ednio da fun\u00e7\u00e3o inversa \u00e9: \\[\\begin{array}{*{35}{l}}
\n{{D}_{{{h}^{-1}}}}=D{{‘}_{h}} & = & \\left\\{ x\\in \\mathbb{R}:\\frac{4-x}{3}>0 \\right\\}\u00a0 \\\\
\n{} & = & \\left] -\\infty ,4 \\right[\u00a0 \\\\
\n\\end{array}\\]<\/p>\nPortanto, a fun\u00e7\u00e3o inversa \u00e9 caraterizada por: \\[\\begin{array}{*{35}{l}}
\n{{h}^{-1}}: & \\left] -\\infty ,4 \\right[\\to \\mathbb{R}\u00a0 \\\\
\n{} & x\\to 2-\\ln \\left( \\frac{4-x}{3} \\right)\u00a0 \\\\
\n\\end{array}\\]![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/01\/12p55-28c.jpg\" "\\"Fun\u00e7\u00e3o")
<\/a><\/p>\n<< R2<\/a><\/span>R4 >><\/a><\/span><\/div><\/div>\n\n\nR4<\/b><\/span><\/p>\n- \n
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$j:x\\to 4-\\ln (1-2x)$<\/p><\/blockquote>\n<\/li>\n<\/ul>\n
Comecemos por determinar o dom\u00ednio da fun\u00e7\u00e3o:<\/p>\n
\\[\\begin{array}{*{35}{l}}
\n{{D}_{j}}=D{{‘}_{{{j}^{-1}}}} & = & \\left\\{ x\\in \\mathbb{R}:1-2x>0 \\right\\}\u00a0 \\\\
\n{} & = & \\left\\{ x\\in \\mathbb{R}:x<\\frac{1}{2} \\right\\}\u00a0 \\\\
\n{} & = & \\left] -\\infty ,\\frac{1}{2} \\right[\u00a0 \\\\
\n\\end{array}\\]<\/p>\n\u00a0Ora, \\[\\begin{array}{*{35}{l}}
\ny=4-\\ln (1-2x) & \\Leftrightarrow\u00a0 & \\ln (1-2x)=4-y\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & 1-2x={{e}^{4-y}}\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & x=\\frac{1-{{e}^{4-y}}}{2}\u00a0 \\\\
\n\\end{array}\\]<\/p>\nLogo, o dom\u00ednio da fun\u00e7\u00e3o inversa \u00e9: \\[\\begin{array}{*{35}{l}}
\n{{D}_{{{j}^{-1}}}}=D{{‘}_{j}} & = & \\left\\{ x\\in \\mathbb{R}:(4-x)\\in \\mathbb{R} \\right\\}\u00a0 \\\\
\n{} & = & \\mathbb{R}\u00a0 \\\\
\n\\end{array}\\]<\/p>\nPortanto, a fun\u00e7\u00e3o inversa \u00e9 caraterizada por: \\[\\begin{array}{*{35}{l}}
\n{{j}^{-1}}: & \\mathbb{R}\\to \\left] -\\infty ,\\frac{1}{2} \\right[\u00a0 \\\\
\n{} & x\\to \\frac{1-{{e}^{4-x}}}{2}\u00a0 \\\\
\n\\end{array}\\]![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/01\/12p55-28d.jpg\" "\\"Fun\u00e7\u00e3o")
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