Sendo f e g funções reais de variável real
Funções com radicais: Infinito 11 A - Parte 2 Pág. 206 Ex. 85
Sendo f e g funções reais de variável real, caracterize $f\circ g$ e $g\circ f$, em cada um dos casos:
- $\begin{matrix}
f(x)=\sqrt{x} & \text{e} & g(x)={{x}^{2}}+1 \\
\end{matrix}$ - $\begin{matrix}
f(x)={{(x-1)}^{3}} & \text{e} & g(x)=\sqrt[3]{x}+1 \\
\end{matrix}$
- Ora, ${{D}_{f\circ g}}=\left\{ x\in \mathbb{R}:x\in {{D}_{g}}\wedge g(x)\in {{D}_{f}} \right\}=\left\{ x\in \mathbb{R}:x\in \mathbb{R}\wedge ({{x}^{2}}+1)\in \mathbb{R}_{0}^{+} \right\}=\mathbb{R}$.
Como \[\begin{array}{*{35}{l}}
(f\circ g)(x) & = & f(g(x)) \\
{} & = & f({{x}^{2}}+1) \\
{} & = & \sqrt{{{x}^{2}}+1} \\
\end{array}\]
então \[\begin{array}{*{35}{l}}
f\circ g: & \mathbb{R}\to \mathbb{R} \\
{} & x\to \sqrt{{{x}^{2}}+1} \\
\end{array}\]Ora, ${{D}_{g\circ f}}=\left\{ x\in \mathbb{R}:x\in {{D}_{f}}\wedge f(x)\in {{D}_{g}} \right\}=\left\{ x\in \mathbb{R}:x\in \mathbb{R}_{0}^{+}\wedge (\sqrt{x})\in \mathbb{R} \right\}=\mathbb{R}_{0}^{+}$.
Como \[\begin{array}{*{35}{l}}
(g\circ f)(x) & = & g(f(x)) \\
{} & = & g(\sqrt{x}) \\
{} & = & \left| x \right|+1 \\
{} & = & x+1\,\,(\text{pois }x\in \mathbb{R}_{0}^{+}) \\
\end{array}\]
então \[\begin{array}{*{35}{l}}
g\circ f: & \mathbb{R}_{0}^{+}\to \mathbb{R} \\
{} & x\to x+1 \\
\end{array}\]
- Ora, ${{D}_{f\circ g}}=\left\{ x\in \mathbb{R}:x\in {{D}_{g}}\wedge g(x)\in {{D}_{f}} \right\}=\left\{ x\in \mathbb{R}:x\in \mathbb{R}\wedge (\sqrt[3]{x}+1)\in \mathbb{R} \right\}=\mathbb{R}$.
Como \[\begin{array}{*{35}{l}}
(f\circ g)(x) & = & f(g(x)) \\
{} & = & f(\sqrt[3]{x}+1) \\
{} & = & {{(\sqrt[3]{x}+1-1)}^{3}} \\
{} & = & x \\
\end{array}\]
então \[\begin{array}{*{35}{l}}
f\circ g: & \mathbb{R}\to \mathbb{R} \\
{} & x\to x \\
\end{array}\]Ora, ${{D}_{g\circ f}}=\left\{ x\in \mathbb{R}:x\in {{D}_{f}}\wedge f(x)\in {{D}_{g}} \right\}=\left\{ x\in \mathbb{R}:x\in \mathbb{R}\wedge ({{(x-1)}^{3}})\in \mathbb{R} \right\}=\mathbb{R}$.
Como \[\begin{array}{*{35}{l}}
(g\circ f)(x) & = & g(f(x)) \\
{} & = & g({{(x-1)}^{3}}) \\
{} & = & \sqrt[3]{{{(x-1)}^{3}}}+1 \\
{} & = & x-1+1 \\
{} & = & x \\
\end{array}\]
então \[\begin{array}{*{35}{l}}
g\circ f: & \mathbb{R}\to \mathbb{R} \\
{} & x\to x \\
\end{array}\]
