Categoria: Função composta

Considere as funções $f$ e $g$ 0

Considere as funções $f$ e $g$

Mais funções: Aleph 11 - Volume 2 Pág. 138 Ex. 10

Enunciado

Considere as funções $f$ e $g$ definidas por: \[\begin{array}{*{20}{c}}
  {f\left( x \right) = \frac{2}{{x – 1}}}&{\text{e}}&{g\left( x \right) = \frac{1}{{x + 3}}}
\end{array}\]

  1. Determine o domínio de cada uma delas.
     
  2. Caracterize as funções $f \circ g$, $g \circ f$ e $f \circ f$.

Resolução >> Resolução

\[\begin{array}{*{20}{c}}
  {f\left( x \right) = \frac{2}{{x – 1}}}&{\text{e}}&{g\left( x \right) = \frac{1}{{x + 3}}}
\end{array}\]

 

  1. ${D_f} = \left\{ {x \in \mathbb{R}:x – 1 \ne 0} \right\} = \mathbb{R}\backslash \left\{ 1 \right\}$.
Caracterize as funções 0

Caracterize as funções

Mais funções: Aleph 11 - Volume 2 Pág. 138 Ex. 8

Enunciado

Considere as funções definidas por:

 \[\begin{array}{*{20}{l}}
  {\begin{array}{*{20}{l}}
  {f:}&{\mathbb{R}\backslash \left\{ 0 \right\} \to \mathbb{R}} \\
  {}&{x \to \frac{1}{{{x^2}}}}
\end{array}}&{}&{}&{\begin{array}{*{20}{l}}
  {g:}&{\mathbb{R}\backslash \left\{ { – 1} \right\} \to \mathbb{R}} \\
  {}&{x \to x + 1}
\end{array}}&{}&{}&{\begin{array}{*{20}{l}}
  {h:}&{\mathbb{R}\backslash \left\{ {0,1} \right\} \to \mathbb{R}} \\
  {}&{x \to \frac{1}{{{x^2} – x}}}
\end{array}}
\end{array}\]

Caracterize as funções seguintes:

\[\begin{array}{*{20}{l}}
  {f + g}&{}&{f \times g}&{}&{\frac{f}{g}}&{}&{h – g}&{}&{\frac{f}{h}}&{}&{f \circ g}&{}&{g \circ f}
\end{array}\]

 

Resolução >> Resolução

\[\begin{array}{*{20}{l}}
  {\begin{array}{*{20}{l}}
  {f:}&{\mathbb{R}\backslash \left\{ 0 \right\} \to \mathbb{R}} \\
  {}&{x \to \frac{1}{{{x^2}}}}
\end{array}}&{}&{}&{\begin{array}{*{20}{l}}
  {g:}&{\mathbb{R}\backslash \left\{ { – 1} \right\} \to \mathbb{R}} \\
  {}&{x \to x + 1}
\end{array}}&{}&{}&{\begin{array}{*{20}{l}}
  {h:}&{\mathbb{R}\backslash \left\{ {0,1} \right\} \to \mathbb{R}} \\
  {}&{x \to \frac{1}{{{x^2} – x}}}
\end{array}}
\end{array}\]


\[{f + g}\]

 

\[{D_{f + g}} = {D_f} \cap {D_g} = \mathbb{R}\backslash \left\{ 0 \right\} \cap \mathbb{R}\backslash \left\{ { – 1} \right\} = \mathbb{R}\backslash \left\{ { – 1,0} \right\}\]

\[\left( {f + g} \right)\left( x \right) = f\left( x \right) + g\left( x \right) = \frac{1}{{{x^2}}} + x + 1 = \frac{{{x^3} + {x^2} + 1}}{{{x^2}}},\forall x \in {D_{f + g}}\]

\[\begin{array}{*{20}{l}}
  {f + g:}&{\mathbb{R}\backslash \left\{ { – 1,0} \right\} \to \mathbb{R}} \\
  {}&{x \to \frac{{{x^3} + {x^2} + 1}}{{{x^2}}}}
\end{array}\]


\[{f \times g}\]

 

\[{D_{f \times g}} = {D_f} \cap {D_g} = \mathbb{R}\backslash \left\{ 0 \right\} \cap \mathbb{R}\backslash \left\{ { – 1} \right\} = \mathbb{R}\backslash \left\{ { – 1,0} \right\}\]

\[\left( {f \times g} \right)\left( x \right) = f\left( x \right) \times g\left( x \right) = \frac{1}{{{x^2}}} \times \left( {x + 1} \right) = \frac{{x + 1}}{{{x^2}}},\forall x \in {D_{f \times g}}\]

\[\begin{array}{*{20}{l}}
  {f \times g:}&{\mathbb{R}\backslash \left\{ { – 1,0} \right\} \to \mathbb{R}} \\
  {}&{x \to \frac{{x + 1}}{{{x^2}}}}
\end{array}\]


\[{\frac{f}{g}}\]

 

\[{D_{\frac{f}{g}}} = {D_f} \cap {D_g} \cap \left\{ {x \in \mathbb{R}:g\left( x \right) \ne 0} \right\} = \mathbb{R}\backslash \left\{ 0 \right\} \cap \mathbb{R}\backslash \left\{ { – 1} \right\} \cap \mathbb{R}\backslash \left\{ { – 1} \right\} = \mathbb{R}\backslash \left\{ { – 1,0} \right\}\]

\[\left( {\frac{f}{g}} \right)\left( x \right) = \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\frac{1}{{{x^2}}}}}{{x + 1}} = \frac{1}{{{x^3} + {x^2}}},\forall x \in {D_{\frac{f}{g}}}\]

\[\begin{array}{*{20}{l}}
  {\frac{f}{g}:}&{\mathbb{R}\backslash \left\{ { – 1,0} \right\} \to \mathbb{R}} \\
  {}&{x \to \frac{1}{{{x^3} + {x^2}}}}
\end{array}\]

 


\[{h – g}\]

 

\[{D_{h – g}} = {D_h} \cap {D_g} = \mathbb{R}\backslash \left\{ {0,1} \right\} \cap \mathbb{R}\backslash \left\{ { – 1} \right\} = \mathbb{R}\backslash \left\{ { – 1,0,1} \right\}\]

\[\left( {h – g} \right)\left( x \right) = h\left( x \right) – g\left( x \right) = \frac{1}{{{x^2} – x}} – \left( {x + 1} \right) = \frac{{ – {x^3} – {x^2} + {x^2} + x + 1}}{{{x^2} – x}} = \frac{{ – {x^3} + x + 1}}{{{x^2} – x}},\forall x \in {D_{\frac{f}{g}}}\]

\[\begin{array}{*{20}{l}}
  {h – g:}&{\mathbb{R}\backslash \left\{ { – 1,0,1} \right\} \to \mathbb{R}} \\
  {}&{x \to \frac{{ – {x^3} + x + 1}}{{{x^2} – x}}}
\end{array}\]

 


\[\frac{f}{h}\]

 

\[{D_{\frac{f}{h}}} = {D_f} \cap {D_h} \cap \left\{ {x \in \mathbb{R}:h\left( x \right) \ne 0} \right\} = \mathbb{R}\backslash \left\{ 0 \right\} \cap \mathbb{R}\backslash \left\{ {0,1} \right\} \cap \mathbb{R}\backslash \left\{ {0,1} \right\} = \mathbb{R}\backslash \left\{ {0,1} \right\}\]

\[\left( {\frac{f}{h}} \right)\left( x \right) = \frac{{f\left( x \right)}}{{h\left( x \right)}} = \frac{{\frac{1}{{{x^2}}}}}{{\frac{1}{{{x^2} – x}}}} = \frac{{{x^2} – x}}{{{x^2}}} = \frac{{x\left( {x – 1} \right)}}{{{x^2}}} = \frac{{x – 1}}{x},\forall x \in {D_{\frac{f}{g}}}\]

\[\begin{array}{*{20}{l}}
  {\frac{f}{h}:}&{\mathbb{R}\backslash \left\{ {0,1} \right\} \to \mathbb{R}} \\
  {}&{x \to \frac{{x – 1}}{x}}
\end{array}\]

 


\[f \circ g\]

 

\[{D_{f \circ g}} = \left\{ {x \in \mathbb{R}:x \in {D_g} \wedge g\left( x \right) \in {D_f}} \right\} = \left\{ {x \in \mathbb{R}:x \in \mathbb{R}\backslash \left\{ { – 1} \right\} \wedge \left( {x + 1} \right) \in \mathbb{R}\backslash \left\{ 0 \right\}} \right\} = \left\{ {x \in \mathbb{R}:x \in \mathbb{R}\backslash \left\{ { – 1} \right\} \wedge x \in \mathbb{R}\backslash \left\{ { – 1} \right\}} \right\} = \mathbb{R}\backslash \left\{ { – 1} \right\}\]

\[\left( {f \circ g} \right)\left( x \right) = f\left( {g\left( x \right)} \right) = f\left( {x + 1} \right) = \frac{1}{{{{\left( {x + 1} \right)}^2}}},\forall x \in {D_{f \circ g}}\]

\[\begin{array}{*{20}{l}}
  {f \circ g:}&{\mathbb{R}\backslash \left\{ { – 1} \right\} \to \mathbb{R}} \\
  {}&{x \to \frac{1}{{{{\left( {x + 1} \right)}^2}}}}
\end{array}\]

 


\[g \circ f\]

 

\[{D_{g \circ f}} = \left\{ {x \in \mathbb{R}:x \in {D_f} \wedge f\left( x \right) \in {D_g}} \right\} = \left\{ {x \in \mathbb{R}:x \in \mathbb{R}\backslash \left\{ 0 \right\} \wedge \left( {\frac{1}{{{x^2}}}} \right) \in \mathbb{R}\backslash \left\{ { – 1} \right\}} \right\} = \left\{ {x \in \mathbb{R}:x \in \mathbb{R}\backslash \left\{ 0 \right\} \wedge x \in \mathbb{R}\backslash \left\{ 0 \right\}} \right\} = \mathbb{R}\backslash \left\{ 0 \right\}\]

\[\left( {g \circ f} \right)\left( x \right) = g\left( {f\left( x \right)} \right) = g\left( {\frac{1}{{{x^2}}}} \right) = \frac{1}{{{x^2}}} + 1 = \frac{{{x^2} + 1}}{{{x^2}}},\forall x \in {D_{f \circ g}}\]

\[\begin{array}{*{20}{l}}
  {g \circ f:}&{\mathbb{R}\backslash \left\{ 0 \right\} \to \mathbb{R}} \\
  {}&{x \to \frac{{{x^2} + 1}}{{{x^2}}}}
\end{array}\]

<< Enunciado
Dadas as funções $f$ e $g$ 0

Dadas as funções $f$ e $g$

Mais funções: Aleph 11 - Volume 2 Pág. 137 Ex. 7

Enunciado

Dadas as funções $f$ e $g$ definidas por: \[\begin{array}{*{20}{c}}
  {f\left( x \right) = 2x + 3}&{\text{e}}&{g\left( x \right) =  – {x^2} + 5}
\end{array}\] determine:

  1. $\left( {f \circ f} \right)\left( 1 \right)$
     
  2. $\left( {g \circ g} \right)\left( 2 \right)$
     
  3. $\left( {f \circ g} \right)\left( 2 \right)$
     
  4. $\left( {g \circ f} \right)\left( 2 \right)$

Resolução >> Resolução

\[\begin{array}{*{20}{c}}   {f\left( x \right) = 2x + 3}&{\text{e}}&{g\left( x \right) =  – {x^2} + 5} \end{array}\]

 

  1. $\left( {f \circ f} \right)\left( 1 \right) = f\left( {f\left( 1 \right)} \right) = f\left( {2 \times 1 + 3} \right) = f\left( 5 \right) = 2 \times 5 + 3 = 13$
      
  2. $\left( {g \circ g} \right)\left( 2 \right) = g\left( {g\left( 2 \right)} \right) = g\left( { – {2^2} + 5} \right) = g\left( 1 \right) =  – {1^2} + 5 = 4$
      
  3. $\left( {f \circ g} \right)\left( 2 \right) = f\left( {g\left( 2 \right)} \right) = f\left( { – {2^2} + 5} \right) = f\left( 1 \right) = 2 \times 1 + 3 = 5$
      
  4. $\left( {g \circ f} \right)\left( 2 \right) = g\left( {f\left( 2 \right)} \right) = g\left( {2 \times 2 + 3} \right) = g\left( 7 \right) =  – {7^2} + 5 =  – 44$

 

<< Enunciado
0

Gráfico de uma função $g$

Mais funções: Aleph 11 - Volume 2 Pág. 131 Ex. 13

Enunciado

Na figura está representado parte do gráfico de uma função quadrática $g$.

Seja $f$ a função, de domínio $\mathbb{R}$, definida por $f\left( x \right) = \left| x \right|$.

Qual é o valor de $\left( {f \circ g} \right)\left( 3 \right)$?

Qual é o valor de $\left( {g \circ f} \right)\left( 3 \right)$?…

Dadas as funções 0

Dadas as funções

Função composta: Infinito 11 A - Parte 2 Pág. 203 Ex. 71

Enunciado

Dadas as funções definidas em $\mathbb{R}$ por \[\begin{matrix}
   f(x)=3x-4 & e & g(x)=\frac{1}{x}  \\
\end{matrix}\]

  1. Determine:
    $(f+g)(5)$ $(f-g)(5)$ $(f\times g)(5)$ $(f\div g)(5)$
    $(f\circ g)(5)$ $(g\circ f)(5)$ $(f\circ f)(5)$ $(g\circ g)(5)$

     

  2. 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 >> Resolução

  1.  
    \[(f+g)(5)=f(5)+g(5)=11+\frac{1}{5}=\frac{56}{5}\]
    \[(f-g)(5)=f(5)-g(5)=11-\frac{1}{5}=\frac{54}{5}\]
    \[(f\times g)(5)=f(5)\times g(5)=11\times \frac{1}{5}=\frac{11}{5}\]
    \[(f\div g)(5)=f(5)\div g(5)=11\div \frac{1}{5}=55\]
    \[(f\circ g)(5)=f(g(5))=f(\frac{1}{5})=3\times \frac{1}{5}-4=-\frac{17}{5}\]
    \[(g\circ f)(5)=g(f(5))=g(11)=\frac{1}{11}\]
    \[(f\circ f)(5)=f(f(5))=f(11)=29\]
    \[(g\circ g)(5)=g(g(5))=g(\frac{1}{5})=5\]
     
  2. Ora, ${{D}_{f}}=\mathbb{R}$ e ${{D}_{g}}=\mathbb{R}\backslash \left\{ 0 \right\}$.
Sendo $f$ e $g$ funções reais de variável real 0

Sendo $f$ e $g$ funções reais de variável real

Função composta: Infinito 11 A - Parte 2 Pág. 202 Ex. 66

Enunciado

Sendo $f$ e $g$ funções reais de variável real, caracterize $f\circ g$ e $g\circ f$ em cada um dos casos:

  1. $\begin{matrix}
       f(x)={{x}^{2}}+2x+1 & e & g(x)=3{{x}^{2}}+1  \\
    \end{matrix}$
     
  2. $\begin{matrix}
       f(x)={{x}^{2}}+2x & e & g(x)=\left| x \right|+1  \\
    \end{matrix}$
     
  3. $\begin{matrix}
       f(x)={{x}^{3}} & e & g(x)=\frac{1}{x-3}  \\
    \end{matrix}$

Resolução >> Resolução

  1. Ora, ${{D}_{f\circ g}}=\left\{ x\in \mathbb{R}:x\in {{D}_{g}}\wedge g(x)\in {{D}_{f}} \right\}=\left\{ x\in \mathbb{R}:x\in \mathbb{R}\wedge (3{{x}^{2}}+1)\in \mathbb{R} \right\}=\mathbb{R}$.
Considere a função real de variável real 0

Considere a função real de variável real

Função composta: Infinito 11 A - Parte 2 Pág. 202 Ex. 65

Enunciado

Considere a função real de variável real assim definida: \[f(x)=5x+3\]

Mostre que as funções $f\circ f$ e ${{f}^{2}}$ são distintas.

(${{f}^{2}}$ designa a função $f\times f$,produto de $f$ por si própria.)

Resolução >> Resolução

Ora, ${{D}_{f\circ f}}=\left\{ x\in \mathbb{R}:x\in {{D}_{f}}\wedge f(x)\in {{D}_{f}} \right\}=\left\{ x\in \mathbb{R}:x\in \mathbb{R}\wedge (5x+3)\in \mathbb{R} \right\}=\mathbb{R}$.…