Resolu\u00e7\u00e3o<\/b><\/span><\/p>\n\\[\\begin{array}{*{20}{c}}
\n{f\\left( x \\right) = 6x + 5}&{}&{}&{g\\left( x \\right) =\u00a0 – \\frac{{12}}{{x + 3}}}
\n\\end{array}\\]<\/p>\n
Como ${D_f} = D{‘_f} = \\mathbb{R}$ e $y = 6x + 5 \\Leftrightarrow x = \\frac{{y – 5}}{6}$, ent\u00e3o: \\[\\begin{array}{*{20}{l}}
\n{{f^{ – 1}}:}&{\\mathbb{R} \\to \\mathbb{R}} \\\\
\n{}&{x \\to \\frac{x}{6} – \\frac{5}{6}}
\n\\end{array}\\]<\/p>\n
Como ${D_g} = \\mathbb{R}\\backslash \\left\\{ { – 3} \\right\\}$, $D{‘_g} = \\mathbb{R}\\backslash \\left\\{ 0 \\right\\}$ e $y =\u00a0 – \\frac{{12}}{{x + 3}} \\Leftrightarrow x =\u00a0 – 3 – \\frac{{12}}{y}$, ent\u00e3o: \\[\\begin{array}{*{20}{l}}
\n{{g^{ – 1}}:}&{\\mathbb{R}\\backslash \\left\\{ 0 \\right\\} \\to \\mathbb{R}} \\\\
\n{}&{x \\to\u00a0 – 3 – \\frac{{12}}{x}}
\n\\end{array}\\]<\/p>\n<\/p>\n