Sejam as funções $f$ e $g$
Operações com funções: Infinito 11 A - Parte 2 Pág. 200 Ex. 57
Enunciado
Sejam \[\begin{matrix}
f:x\to \frac{2x+2}{{{x}^{2}}-3x+2} & e & g:x\to \frac{4x-4}{x-2} \\
\end{matrix}\]
- Mostre que $f\times g$ e $\frac{f}{g}$ são funções racionais e determine o seu domínio.
- Determine os valores de x tais que $f(x)\le \frac{1}{2}$.
Resolução
- ${{D}_{f}}=\left\{ x\in \mathbb{R}:{{x}^{2}}-3x+2\ne 0 \right\}=\left\{ x\in \mathbb{R}:\tilde{\ }\left( x=\frac{3\pm \sqrt{9-8}}{2} \right) \right\}=\mathbb{R}\backslash \left\{ 1,2 \right\}$
${{D}_{g}}=\left\{ x\in \mathbb{R}:x-2\ne 0 \right\}=\mathbb{R}\backslash \left\{ 2 \right\}$
${{D}_{f\times g}}={{D}_{f}}\cap {{D}_{g}}=\mathbb{R}\backslash \left\{ 1,2 \right\}$
${{D}_{\frac{f}{g}}}={{D}_{f}}\cap {{D}_{g}}\cap \left\{ x\in \mathbb{R}:g(x)\ne 0 \right\}={{D}_{f}}\cap {{D}_{g}}\cap \left\{ x\in \mathbb{R}:x\ne 1 \right\}=\mathbb{R}\backslash \left\{ 1,2 \right\}$
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, \[\begin{array}{*{35}{l}}
\frac{f}{g}: & \mathbb{R}\backslash \left\{ 1,2 \right\}\to \mathbb{R} \\
{} & x\to \frac{x+1}{2{{x}^{2}}-4x+2} \\
\end{array}\]
- 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, \[f(x)\le \frac{1}{2}\Leftrightarrow x\in \left] -\infty ,\frac{7-\sqrt{57}}{2} \right]\cup \left] 1,2 \right[\cup \left[ \frac{7+\sqrt{57}}{2},+\infty \right[\]
