Sejam as funções $f$ e $g$

Sejam \[\begin{matrix}
f:x\to \frac{2x+2}{{{x}^{2}}-3x+2} & e & g:x\to \frac{4x-4}{x-2}  \\
\end{matrix}\]

­

1. ${{D}_{f}}=\left\{ x\in \mathbb{R}:{{x}^{2}}-3x+2\ne 0 \right\}=\left\{ x\in \mathbb{R}:\tilde{\ }\left( x=\frac{3\pm \sqrt{9-8}}{2} \right) \right\}=\mathbb{R}\backslash \left\{ 1,2 \right\}$

   ${{D}_{g}}=\left\{ x\in \mathbb{R}:x-2\ne 0 \right\}=\mathbb{R}\backslash \left\{ 2 \right\}$

   ${{D}_{f\times g}}={{D}_{f}}\cap {{D}_{g}}=\mathbb{R}\backslash \left\{ 1,2 \right\}$

   ${{D}_{\frac{f}{g}}}={{D}_{f}}\cap {{D}_{g}}\cap \left\{ x\in \mathbb{R}:g(x)\ne 0 \right\}={{D}_{f}}\cap {{D}_{g}}\cap \left\{ x\in \mathbb{R}:x\ne 1 \right\}=\mathbb{R}\backslash \left\{ 1,2 \right\}$

   Ora,
   \[\begin{array}{*{35}{l}}
   (f\times g)(x) & = & f(x)\times g(x)  \\
   {} & = & \frac{2x+2}{{{x}^{2}}-3x+2}\times \frac{4x-4}{x-2}  \\
   {} & = & \frac{2(x+1)}{(x-1)(x-2)}\times \frac{4(x-1)}{x-2}  \\
   {} & = & \frac{2(x+1)}{(x-2)}\times \frac{4}{x-2}  \\
   {} & = & \frac{8x+8}{{{x}^{2}}-4x+4}  \\
   \end{array}\]

   Logo,
   \[\begin{array}{*{35}{l}}
   f\times g: & \mathbb{R}\backslash \left\{ 1,2 \right\}\to \mathbb{R}  \\
   {} & x\to \frac{8x+8}{{{x}^{2}}-4x+4}  \\
   \end{array}\]

   Ora,
   \[\begin{array}{*{35}{l}}
   (\frac{f}{g})(x) & = & \frac{f(x)}{g(x)}  \\
   {} & = & \frac{2x+2}{{{x}^{2}}-3x+2}\div \frac{4x-4}{x-2}  \\
   {} & = & \frac{2(x+1)}{(x-1)(x-2)}\times \frac{x-2}{4(x-1)}  \\
   {} & = & \frac{(x+1)}{(x-1)}\times \frac{1}{2(x-1)}  \\
   {} & = & \frac{x+1}{2{{x}^{2}}-4x+2}  \\
   \end{array}\]

   Logo,
   \[\begin{array}{*{35}{l}}
   \frac{f}{g}: & \mathbb{R}\backslash \left\{ 1,2 \right\}\to \mathbb{R}  \\
   {} & x\to \frac{x+1}{2{{x}^{2}}-4x+2}  \\
   \end{array}\]
   ­

2. Ora,
   \[\begin{array}{*{35}{l}}
   f(x)\le \frac{1}{2} & \Leftrightarrow  & \frac{2x+2}{{{x}^{2}}-3x+2}\le \frac{1}{2}  \\
   {} & \Leftrightarrow  & \frac{2x+2}{(x-1)(x-2)}-\frac{1}{2}\le 0  \\
   {} & \Leftrightarrow  & \frac{4x+4-{{x}^{2}}+3x-2}{2(x-1)(x-2)}\le 0  \\
   {} & \Leftrightarrow  & \frac{-{{x}^{2}}+7x+2}{2(x-1)(x-2)}\le 0  \\
   \end{array}\]

   Cálculo auxiliar:
   \[ – {x^2} + 7x + 2 = 0 \Leftrightarrow x = \frac{{ – 7 \pm \sqrt {49 + 8} }}{{ – 2}} \Leftrightarrow x = \frac{{ – 7 \pm \sqrt {57} }}{{ – 2}} \Leftrightarrow x = \frac{{7 – \sqrt {57} }}{2} \vee x = \frac{{7 + \sqrt {57} }}{2}\]

   Logo, vem:

   $x$	$-\infty $	$\frac{7-\sqrt{57}}{2}$		$1$		$2$		$\frac{7+\sqrt{57}}{2}$	$+\infty $
   $-{{x}^{2}}+7x+2$	–	$0$	+	+	+	+	+	$0$	–
   $2(x-1)(x-2)$	+	+	+	$0$	–	$0$	+	+	+
   $\frac{-{{x}^{2}}+7x+2}{2(x-1)(x-2)}$	–	$0$	+	n.d.	–	n.d.	+	$0$	–

   Portanto, \[f(x)\le \frac{1}{2}\Leftrightarrow x\in \left] -\infty ,\frac{7-\sqrt{57}}{2} \right]\cup \left] 1,2 \right[\cup \left[ \frac{7+\sqrt{57}}{2},+\infty  \right[\]