Sejam as funções racionais
Operações com funções: Infinito 11 A - Parte 2 Pág. 200 Ex. 55
Enunciado
Sejam as funções racionais definidas por: \[\begin{matrix}
f(x)=\frac{1}{4x+3} & e & g(x)=\frac{2x-1}{(4x+3)(x-7)} \\
\end{matrix}\]
- Indique o seu domínio.
- Caracterize $f+g$.
- Determine $x\in \mathbb{R}$ tal que $f(x)\le g(x)$.
Resolução
- ${{D}_{f}}=\left\{ x\in \mathbb{R}:4x+3\ne 0 \right\}=\mathbb{R}\backslash \left\{ -\frac{3}{4} \right\}$
${{D}_{g}}=\left\{ x\in \mathbb{R}:(4x+3)(x-7)\ne 0 \right\}=\mathbb{R}\backslash \left\{ -\frac{3}{4},7 \right\}$
- ${{D}_{f+g}}={{D}_{f}}\cap {{D}_{g}}=\mathbb{R}\backslash \left\{ -\frac{3}{4},7 \right\}$\[\begin{array}{*{35}{l}}
(f+g)(x) & = & f(x)+g(x) \\
{} & = & \frac{1}{4x+3}+\frac{2x-1}{(4x+3)(x-7)} \\
{} & = & \frac{(x-7)+2x-1}{(4x+3)(x-7)} \\
{} & = & \frac{3x-8}{4{{x}^{2}}-25x-21} \\
\end{array}\]Logo,
\[\begin{array}{*{35}{l}}
f+g: & \mathbb{R}\backslash \left\{ -\frac{3}{4},7 \right\}\to \mathbb{R} \\
{} & x\to \frac{3x-8}{4{{x}^{2}}-25x-21} \\
\end{array}\]
- 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, $f(x)\le g(x)\Leftrightarrow x\in \left[ -6,-\frac{3}{4} \right[\cup \left] 7,+\infty \right[$.
