Sejam as funções racionais

Sejam as funções racionais definidas por: \[\begin{matrix}
f(x)=\frac{1}{4x+3} & e & g(x)=\frac{2x-1}{(4x+3)(x-7)}  \\
\end{matrix}\]

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1. ${{D}_{f}}=\left\{ x\in \mathbb{R}:4x+3\ne 0 \right\}=\mathbb{R}\backslash \left\{ -\frac{3}{4} \right\}$

   ${{D}_{g}}=\left\{ x\in \mathbb{R}:(4x+3)(x-7)\ne 0 \right\}=\mathbb{R}\backslash \left\{ -\frac{3}{4},7 \right\}$

2. ${{D}_{f+g}}={{D}_{f}}\cap {{D}_{g}}=\mathbb{R}\backslash \left\{ -\frac{3}{4},7 \right\}$\[\begin{array}{*{35}{l}}
   (f+g)(x) & = & f(x)+g(x)  \\
   {} & = & \frac{1}{4x+3}+\frac{2x-1}{(4x+3)(x-7)}  \\
   {} & = & \frac{(x-7)+2x-1}{(4x+3)(x-7)}  \\
   {} & = & \frac{3x-8}{4{{x}^{2}}-25x-21}  \\
   \end{array}\]Logo,

   \[\begin{array}{*{35}{l}}
   f+g: & \mathbb{R}\backslash \left\{ -\frac{3}{4},7 \right\}\to \mathbb{R}  \\
   {} & x\to \frac{3x-8}{4{{x}^{2}}-25x-21}  \\
   \end{array}\]

3. Ora, \[\begin{array}{*{35}{l}}
   f(x)\le g(x) & \Leftrightarrow  & \frac{1}{4x+3}\le \frac{2x-1}{(4x+3)(x-7)}  \\
   {} & \Leftrightarrow  & \frac{(x-7)-(2x-1)}{(4x+3)(x-7)}\le 0  \\
   {} & \Leftrightarrow  & \frac{-x-6}{(4x+3)(x-7)}\le 0  \\
   {} & \Leftrightarrow  & \frac{x+6}{(4x+3)(x-7)}\ge 0  \\
   \end{array}\]Assim, vem:

   $x$	$-\infty $	$-6$		$-\frac{3}{4}$		$7$	$+\infty $
   $x+6$	–	$0$	+	+	+	+	+
   $(4x+3)(x-7)$	+	+	+	$0$	–	$0$	+
   $\frac{x+6}{(4x+3)(x-7)}$	–	$0$	+	n.d.	–	n.d.	+

   Portanto, $f(x)\le g(x)\Leftrightarrow x\in \left[ -6,-\frac{3}{4} \right[\cup \left] 7,+\infty  \right[$.