Dadas as funções
Função composta: Infinito 11 A - Parte 2 Pág. 203 Ex. 71
Enunciado
Dadas as funções definidas em $\mathbb{R}$ por \[\begin{matrix}
f(x)=3x-4 & e & g(x)=\frac{1}{x} \\
\end{matrix}\]
- Determine:
$(f+g)(5)$ $(f-g)(5)$ $(f\times g)(5)$ $(f\div g)(5)$ $(f\circ g)(5)$ $(g\circ f)(5)$ $(f\circ f)(5)$ $(g\circ g)(5)$ - Caracterize as funções:
$f+g$ $f-g$ $f\times g$ $f\div g$ $f\circ g$ $g\circ f$ $f\circ f$ $g\circ g$
Resolução
- Tem-se sucessivamente:
\[(f+g)(5)=f(5)+g(5)=11+\frac{1}{5}=\frac{56}{5}\]
\[(f-g)(5)=f(5)-g(5)=11-\frac{1}{5}=\frac{54}{5}\]
\[(f\times g)(5)=f(5)\times g(5)=11\times \frac{1}{5}=\frac{11}{5}\]
\[(f\div g)(5)=f(5)\div g(5)=11\div \frac{1}{5}=55\]
\[(f\circ g)(5)=f(g(5))=f(\frac{1}{5})=3\times \frac{1}{5}-4=-\frac{17}{5}\]
\[(g\circ f)(5)=g(f(5))=g(11)=\frac{1}{11}\]
\[(f\circ f)(5)=f(f(5))=f(11)=29\]
\[(g\circ g)(5)=g(g(5))=g(\frac{1}{5})=5\] - Ora, ${{D}_{f}}=\mathbb{R}$ e ${{D}_{g}}=\mathbb{R}\backslash \left\{ 0 \right\}$.
Logo, ${{D}_{f+g}}={{D}_{f}}\cap {{D}_{g}}=\mathbb{R}\backslash \left\{ 0 \right\}$.
Como $(f+g)(x)=f(x)+g(x)=3x-4+\frac{1}{x}$, então \[\begin{array}{*{35}{l}}
f+g: & \mathbb{R}\backslash \left\{ 0 \right\}\to \mathbb{R} \\
{} & x\to 3x-4+\frac{1}{x} \\
\end{array}\]
Ora, ${{D}_{f-g}}={{D}_{f}}\cap {{D}_{g}}=\mathbb{R}\backslash \left\{ 0 \right\}$.
Como $(f-g)(x)=f(x)-g(x)=3x-4-\frac{1}{x}$, então \[\begin{array}{*{35}{l}}
f-g: & \mathbb{R}\backslash \left\{ 0 \right\}\to \mathbb{R} \\
{} & x\to 3x-4-\frac{1}{x} \\
\end{array}\]
Ora, ${{D}_{f\times g}}={{D}_{f}}\cap {{D}_{g}}=\mathbb{R}\backslash \left\{ 0 \right\}$.
Como $(f\times g)(x)=f(x)\times g(x)=\frac{3x-4}{x}$, então \[\begin{array}{*{35}{l}}
f\times g: & \mathbb{R}\backslash \left\{ 0 \right\}\to \mathbb{R} \\
{} & x\to \frac{3x-4}{x} \\
\end{array}\]
Ora, ${{D}_{f\div g}}={{D}_{f}}\cap {{D}_{g}}\cap \left\{ x\in \mathbb{R}:g(x)\ne 0 \right\}=\mathbb{R}\backslash \left\{ 0 \right\}$.
Como $(f\div g)(x)=f(x)\div g(x)=3{{x}^{2}}-4x$, então \[\begin{array}{*{35}{l}}
f\div g: & \mathbb{R}\backslash \left\{ 0 \right\}\to \mathbb{R} \\
{} & x\to 3{{x}^{2}}-4x \\
\end{array}\]
Ora, ${{D}_{f\circ g}}=\left\{ x\in \mathbb{R}:x\in {{D}_{g}}\wedge g(x)\in {{D}_{f}} \right\}=\left\{ x\in \mathbb{R}:x\in \mathbb{R}\backslash \left\{ 0 \right\}\wedge (\frac{1}{x})\in \mathbb{R} \right\}=\mathbb{R}\backslash \left\{ 0 \right\}$.
Como $(f\circ g)(x)=f(g(x))=f(\frac{1}{x})=\frac{3}{x}-4$, então \[\begin{array}{*{35}{l}}
f\circ g: & \mathbb{R}\backslash \left\{ 0 \right\}\to \mathbb{R} \\
{} & x\to \frac{3}{x}-4 \\
\end{array}\]
Ora, ${{D}_{g\circ f}}=\left\{ x\in \mathbb{R}:x\in {{D}_{f}}\wedge f(x)\in {{D}_{g}} \right\}=\left\{ x\in \mathbb{R}:x\in \mathbb{R}\wedge (3x-4)\in \mathbb{R}\backslash \left\{ 0 \right\} \right\}=\mathbb{R}\backslash \left\{ \frac{4}{3} \right\}$.
Como $(g\circ f)(x)=g(f(x))=g(3x-4)=\frac{1}{3x-4}$, então \[\begin{array}{*{35}{l}}
g\circ f: & \mathbb{R}\backslash \left\{ \frac{4}{3} \right\}\to \mathbb{R} \\
{} & x\to \frac{1}{3x-4} \\
\end{array}\]
Ora, ${{D}_{f\circ f}}=\left\{ x\in \mathbb{R}:x\in {{D}_{f}}\wedge f(x)\in {{D}_{f}} \right\}=\left\{ x\in \mathbb{R}:x\in \mathbb{R}\wedge (3x-4)\in \mathbb{R} \right\}=\mathbb{R}$.
Como $(f\circ f)(x)=f(f(x))=f(3x-4)=3(3x-4)-4=9x-16$, então \[\begin{array}{*{35}{l}}
f\circ f: & \mathbb{R}\to \mathbb{R} \\
{} & x\to 9x-16 \\
\end{array}\]
Ora, ${{D}_{g\circ g}}=\left\{ x\in \mathbb{R}:x\in {{D}_{g}}\wedge g(x)\in {{D}_{g}} \right\}=\left\{ x\in \mathbb{R}:x\in \mathbb{R}\backslash \left\{ 0 \right\}\wedge (\frac{1}{x})\in \mathbb{R}\backslash \left\{ 0 \right\} \right\}=\mathbb{R}\backslash \left\{ 0 \right\}$.
Como $(g\circ g)(x)=g(g(x))=g(\frac{1}{x})=x$, então \[\begin{array}{*{35}{l}}
g\circ g: & \mathbb{R}\backslash \left\{ 0 \right\}\to \mathbb{R} \\
{} & x\to x \\
\end{array}\]