Sendo $f$ e $g$ funções reais de variável real
Função composta: Infinito 11 A - Parte 2 Pág. 202 Ex. 66
Sendo $f$ e $g$ funções reais de variável real, caracterize $f\circ g$ e $g\circ f$ em cada um dos casos:
- $\begin{matrix}
f(x)={{x}^{2}}+2x+1 & e & g(x)=3{{x}^{2}}+1 \\
\end{matrix}$ - $\begin{matrix}
f(x)={{x}^{2}}+2x & e & g(x)=\left| x \right|+1 \\
\end{matrix}$ - $\begin{matrix}
f(x)={{x}^{3}} & e & g(x)=\frac{1}{x-3} \\
\end{matrix}$
- Ora, ${{D}_{f\circ g}}=\left\{ x\in \mathbb{R}:x\in {{D}_{g}}\wedge g(x)\in {{D}_{f}} \right\}=\left\{ x\in \mathbb{R}:x\in \mathbb{R}\wedge (3{{x}^{2}}+1)\in \mathbb{R} \right\}=\mathbb{R}$.
Como
\[\begin{array}{*{35}{l}}
(f\circ g)(x) & = & f(g(x)) \\
{} & = & f(3{{x}^{2}}+1) \\
{} & = & {{(3{{x}^{2}}+1)}^{2}}+2(3{{x}^{2}}+1)+1 \\
{} & = & 9{{x}^{4}}+6{{x}^{2}}+1+6{{x}^{2}}+2+1 \\
{} & = & 9{{x}^{4}}+12{{x}^{2}}+4 \\
\end{array}\]
então, \[\begin{array}{*{35}{l}}
f\circ g: & \mathbb{R}\to \mathbb{R} \\
{} & x\to 9{{x}^{4}}+12{{x}^{2}}+4 \\
\end{array}\]Ora, ${{D}_{g\circ f}}=\left\{ x\in \mathbb{R}:x\in {{D}_{f}}\wedge f(x)\in {{D}_{g}} \right\}=\left\{ x\in \mathbb{R}:x\in \mathbb{R}\wedge ({{x}^{2}}+2x+1)\in \mathbb{R} \right\}=\mathbb{R}$.
Como
\[\begin{array}{*{35}{l}}
(g\circ f)(x) & = & g(f(x)) \\
{} & = & g({{x}^{2}}+2x+1) \\
{} & = & 3{{({{x}^{2}}+2x+1)}^{2}}+1 \\
{} & = & 3({{x}^{2}}+2x+1)({{x}^{2}}+2x+1)+1 \\
{} & = & 3({{x}^{4}}+2{{x}^{3}}+{{x}^{2}}+2{{x}^{3}}+4{{x}^{2}}+2x+{{x}^{2}}+2x+1)+1 \\
{} & = & 3{{x}^{4}}+12{{x}^{3}}+18{{x}^{2}}+12x+4 \\
\end{array}\]
então, \[\begin{array}{*{35}{l}}
g\circ f: & \mathbb{R}\to \mathbb{R} \\
{} & x\to 3{{x}^{4}}+12{{x}^{3}}+18{{x}^{2}}+12x+4 \\
\end{array}\] - Ora, ${{D}_{f\circ g}}=\left\{ x\in \mathbb{R}:x\in {{D}_{g}}\wedge g(x)\in {{D}_{f}} \right\}=\left\{ x\in \mathbb{R}:x\in \mathbb{R}\wedge (\left| x \right|+1)\in \mathbb{R} \right\}=\mathbb{R}$.
Como
\[\begin{array}{*{35}{l}}
(f\circ g)(x) & = & f(g(x)) \\
{} & = & f(\left| x \right|+1) \\
{} & = & {{(\left| x \right|+1)}^{2}}+2(\left| x \right|+1) \\
{} & = & {{x}^{2}}+2\left| x \right|+1+2\left| x \right|+2 \\
{} & = & {{x}^{2}}+4\left| x \right|+3 \\
\end{array}\]
então, \[\begin{array}{*{35}{l}}
f\circ g: & \mathbb{R}\to \mathbb{R} \\
{} & x\to {{x}^{2}}+4\left| x \right|+3 \\
\end{array}\]Ora, ${{D}_{g\circ f}}=\left\{ x\in \mathbb{R}:x\in {{D}_{f}}\wedge f(x)\in {{D}_{g}} \right\}=\left\{ x\in \mathbb{R}:x\in \mathbb{R}\wedge ({{x}^{2}}+2x)\in \mathbb{R} \right\}=\mathbb{R}$.
Como
\[\begin{array}{*{35}{l}}
(g\circ f)(x) & = & g(f(x)) \\
{} & = & g({{x}^{2}}+2x) \\
{} & = & \left| {{x}^{2}}+2x \right|+1 \\
\end{array}\]
então, \[\begin{array}{*{35}{l}}
g\circ f: & \mathbb{R}\to \mathbb{R} \\
{} & x\to \left| {{x}^{2}}+2x \right|+1 \\
\end{array}\] - Ora, ${{D}_{f\circ g}}=\left\{ x\in \mathbb{R}:x\in {{D}_{g}}\wedge g(x)\in {{D}_{f}} \right\}=\left\{ x\in \mathbb{R}:x\in \mathbb{R}\backslash \left\{ 3 \right\}\wedge (\frac{1}{x-3})\in \mathbb{R} \right\}=\mathbb{R}\backslash \left\{ 3 \right\}$.
Como
\[\begin{array}{*{35}{l}}
(f\circ g)(x) & = & f(g(x)) \\
{} & = & f(\frac{1}{x-3}) \\
{} & = & {{(\frac{1}{x-3})}^{3}} \\
{} & = & \frac{1}{{{(x-3)}^{3}}} \\
\end{array}\]
então, \[\begin{array}{*{35}{l}}
f\circ g: & \mathbb{R}\backslash \left\{ 3 \right\}\to \mathbb{R} \\
{} & x\to \frac{1}{{{(x-3)}^{3}}} \\
\end{array}\]Ora, ${{D}_{g\circ f}}=\left\{ x\in \mathbb{R}:x\in {{D}_{f}}\wedge f(x)\in {{D}_{g}} \right\}=\left\{ x\in \mathbb{R}:x\in \mathbb{R}\wedge ({{x}^{3}})\in \mathbb{R}\backslash \left\{ 3 \right\} \right\}=\mathbb{R}\backslash \left\{ \sqrt[3]{3} \right\}$.
Como
\[\begin{array}{*{35}{l}}
(g\circ f)(x) & = & g(f(x)) \\
{} & = & g({{x}^{3}}) \\
{} & = & \frac{1}{{{x}^{3}}-3} \\
\end{array}\]
então, \[\begin{array}{*{35}{l}}
g\circ f: & \mathbb{R}\backslash \left\{ \sqrt[3]{3} \right\}\to \mathbb{R} \\
{} & x\to \frac{1}{{{x}^{3}}-3} \\
\end{array}\]
