Categoria: Operações com funções

Mostre que as funções são iguais 0

Mostre que as funções são iguais

Mais funções: Aleph 11 - Volume 2 Pág. 139 Ex. 14

Enunciado

Mostre que as funções seguintes são iguais.

\[\begin{array}{*{20}{c}}
  {\begin{array}{*{20}{c}}
  {f:}&{\mathbb{R}\backslash \left\{ { – 3,3} \right\} \to \mathbb{R}} \\
  {}&{x \to \frac{{x – 3}}{{{x^2} – 9}}}
\end{array}}&{}&{\text{e}}&{}&{\begin{array}{*{20}{c}}
  {g:}&{\mathbb{R}\backslash \left\{ { – 3,3} \right\} \to \mathbb{R}} \\
  {}&{x \to \frac{1}{{x + 3}}}
\end{array}}
\end{array}\]

Resolução >> Resolução

\[\begin{array}{*{20}{c}}
  {\begin{array}{*{20}{c}}
  {f:}&{\mathbb{R}\backslash \left\{ { – 3,3} \right\} \to \mathbb{R}} \\
  {}&{x \to \frac{{x – 3}}{{{x^2} – 9}}}
\end{array}}&{}&{\text{e}}&{}&{\begin{array}{*{20}{c}}
  {g:}&{\mathbb{R}\backslash \left\{ { – 3,3} \right\} \to \mathbb{R}} \\
  {}&{x \to \frac{1}{{x + 3}}}
\end{array}}
\end{array}\]

Como \[{D_f} = \mathbb{R}\backslash \left\{ { – 3,3} \right\} = {D_g}\] e \[f\left( x \right) = \frac{{x – 3}}{{{x^2} – 9}} = \frac{{x – 3}}{{\left( {x + 3} \right)\left( {x – 3} \right)}} = \frac{1}{{x + 3}} = g\left( x \right),\,\forall x \in {D_f}\] então as funções $f$ e $g$ são iguais.…

As funções de Heaviside e rampa 0

As funções de Heaviside e rampa

Mais funções: Aleph 11 - Volume 2 Pág. 139 Ex. 12

Enunciado

 As funções de Heaviside e rampa são definidas, respetivamente, por: \[\begin{array}{*{20}{c}}
  {H\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
  0& \Leftarrow &{x < 0} \\
  {\frac{1}{2}}& \Leftarrow &{x = 0} \\
  1& \Leftarrow &{x > 0}
\end{array}} \right.}&{\text{e}}&{R\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
  0& \Leftarrow &{x \leqslant 0} \\
  x& \Leftarrow &{x > 0}
\end{array}} \right.}
\end{array}\]

Mostre que:

  1. $R\left( x \right) = x\,H\left( x \right)$
     
  2. $R\left( x \right) = \frac{{x + \left| x \right|}}{2}$
     
  3. $\left( {R \circ R} \right)\left( x \right) = R\left( x \right)$

Resolução >> Resolução

\[\begin{array}{*{20}{c}}
  {H\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
  0& \Leftarrow &{x < 0} \\
  {\frac{1}{2}}& \Leftarrow &{x = 0} \\
  1& \Leftarrow &{x > 0}
\end{array}} \right.}&{\text{e}}&{R\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
  0& \Leftarrow &{x \leqslant 0} \\
  x& \Leftarrow &{x > 0}
\end{array}} \right.}
\end{array}\]

 

  1. \[\begin{array}{*{20}{c}}
      {x\,H\left( x \right)}& = &{x \times \left\{ {\begin{array}{*{20}{c}}
      0& \Leftarrow &{x < 0} \\
      {\frac{1}{2}}& \Leftarrow &{x = 0} \\
      1& \Leftarrow &{x > 0}
    \end{array}} \right.}& = &{\left\{ {\begin{array}{*{20}{c}}
      {x \times 0}& \Leftarrow &{x < 0} \\
      {x \times \frac{1}{2}}& \Leftarrow &{x = 0} \\
      {x \times 1}& \Leftarrow &{x > 0}
    \end{array}} \right.}& = &{\left\{ {\begin{array}{*{20}{c}}
      0& \Leftarrow &{x < 0} \\
      0& \Leftarrow &{x = 0} \\
      x& \Leftarrow &{x > 0}
    \end{array}} \right.}& = &{R\left( x \right)}
    \end{array}\]
     
  2. \[\begin{array}{*{20}{c}}
      {\frac{{x + \left| x \right|}}{2}}& = &{\left\{ {\begin{array}{*{20}{c}}
      {\frac{{x + \left( { – x} \right)}}{2}}& \Leftarrow &{x \leqslant 0} \\
      {\frac{{x + x}}{2}}& \Leftarrow &{x > 0}
    \end{array}} \right.}& = &{\left\{ {\begin{array}{*{20}{c}}
      0& \Leftarrow &{x \leqslant 0} \\
      x& \Leftarrow &{x > 0}
    \end{array}} \right.}& = &{R\left( x \right)}
    \end{array}\]
     
  3. \[\begin{array}{*{20}{c}}
      {\left( {R \circ R} \right)\left( x \right)}& = &{R\left( {R\left( x \right)} \right)}& = &{R\left( {\left\{ {\begin{array}{*{20}{c}}
      0& \Leftarrow &{x \leqslant 0} \\
      x& \Leftarrow &{x > 0}
    \end{array}} \right.} \right)}& = &{R\left( x \right)}
    \end{array}\]

 

<< Enunciado
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
0

Uma função quadrática e uma função afim

Mais funções: Aleph 11 - Volume 2 Pág. 127 Ex. 12

Enunciado

Na figura estão representadas:

  • parte do gráfico de uma função quadrática $f$;
     
  • parte do gráfico de uma função afim $g$.

Determine o domínio de cada uma das seguintes funções: $\frac{f}{g}$ e $\frac{g}{f}$.

Resolução >> Resolução

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

\[{D_{\frac{g}{f}}} = {D_g} \cap {D_f} \cap \left\{ {x \in \mathbb{R}:f\left( x \right) \ne 0} \right\} = \mathbb{R} \cap \mathbb{R} \cap \mathbb{R}\backslash \left\{ { – 4,0} \right\} = \mathbb{R}\backslash \left\{ { – 4,0} \right\}\]

<< Enunciado
Considere as funções 0

Considere as funções

Mais funções: Aleph 11 - Volume 2 Pág. 127 Ex. 11

Enunciado

Considere as funções definidas por:

 

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

Caracterize as seguintes funções:

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

Resolução >> Resolução

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

 

\[{f + g}\]

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

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

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

 
\[{f \times g}\]

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

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

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

 

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

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

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

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

 

\[{h – g}\]

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

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

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

 

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

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

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

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

<< Enunciado
0

Três funções: $f$, $g$ e $\frac{f}{g}$

Mais funções: Aleph 11 - Volume 2 Pág. 125 Ex. 10

Enunciado
Sejam $f$ e $g$ duas funções definidas por: \[\begin{array}{*{20}{c}}
  {f\left( x \right) = {x^2} – 4}&{\text{e}}&{g\left( x \right) = x + 2}
\end{array}\]

Caracterize a função $\frac{f}{g}$ e estude o seu sinal, relacionando-o com o sinal quer da função $f$ quer da função $g$.

Resolução >> Resolução

\[\begin{array}{*{20}{c}}
  {f\left( x \right) = {x^2} – 4}&{\text{e}}&{g\left( x \right) = x + 2}
\end{array}\]

Comecemos por determinar o domínio de $\frac{f}{g}$:

\[\begin{array}{*{20}{l}}
  {{D_{\frac{f}{g}}}}& = &{{D_f} \cap {D_g} \cap \left\{ {x \in \mathbb{R}:g\left( x \right) \ne 0} \right\}} \\
  {}& = &{\mathbb{R} \cap \mathbb{R} \cap \left\{ {x \in \mathbb{R}:x + 2 \ne 0} \right\}} \\
  {}& = &{\mathbb{R}\backslash \left\{ { – 2} \right\}}
\end{array}\]

 Determinemos, agora, $\frac{f}{g}\left( x \right)$:

 

\[\frac{f}{g}\left( x \right) = \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{{x^2} – 4}}{{x + 2}} = \frac{{\left( {x + 2} \right)\left( {x – 2} \right)}}{{x + 2}} = x – 2,\,\forall x \in \mathbb{R}\backslash \left\{ { – 2} \right\}\]Logo:

\[\begin{array}{*{20}{r}}
  {\frac{f}{g}:}&{\mathbb{R}\backslash \left\{ { – 2} \right\} \to \mathbb{R}} \\
  {}&{x \to x – 2}
\end{array}\]

 

$x$ $ – \infty $ $-2$   $2$ $ + \infty $
$\frac{f}{g}\left( x \right) = x – 2$ $-$ n.d.
0

Duas funções, $s$ e $t$

Mais funções: Aleph 11 - Volume 2 Pág. 122 Ex. 9

Enunciado

Na figura estão representadas graficamente as funções $s$ e $t$.

Determine:

  1. $s\left( 0 \right)$
     
  2. $t\left( 5 \right)$
     
  3. $\left( {s + t} \right)\left( 3 \right)$
     
  4. $\left( {s – t} \right)\left( 3 \right)$

Resolução >> Resolução

  1. $s\left( 0 \right) = 2$
     
  2. $t\left( 5 \right) = 0$
     
  3. $\left( {s + t} \right)\left( 3 \right) = s\left( 3 \right) + t\left( 3 \right) = 4 + 4 = 8$
     
  4. $\left( {s – t} \right)\left( 3 \right) = s\left( 3 \right) – t\left( 3 \right) = 4 – 4 = 0$

 

<< Enunciado
Verifique se são iguais as funções 0

Verifique se são iguais as funções

Mais funções: Aleph 11 - Volume 2 Pág. 118 Ex. 8

Enunciado

Verifique se são iguais os seguintes pares de funções reais de variável real:

  1. \[\begin{array}{*{20}{l}}
      {f\left( x \right) = \frac{{2 – x}}{{{x^2} – 4}}}&{\text{e}}&{g\left( x \right) = \frac{{ – 1}}{{x + 2}}}
    \end{array}\]
  2. \[\begin{array}{*{20}{l}}
      {f\left( x \right) = \frac{x}{{x – 1}}}&{\text{e}}&{g\left( x \right) = \frac{{{x^2} – x}}{{{{\left( {x – 1} \right)}^2}}}}
    \end{array}\]
  3. \[\begin{array}{*{20}{l}}
      {f\left( x \right) = \frac{{{x^4} – 25}}{{{x^2} + 5}}}&{\text{e}}&{g\left( x \right) = {x^2} – 5}
    \end{array}\]

Resolução >> Resolução

  1. \[\begin{array}{*{20}{l}}   {f\left( x \right) = \frac{{2 – x}}{{{x^2} – 4}}}&{\text{e}}&{g\left( x \right) = \frac{{ – 1}}{{x + 2}}} \end{array}\]
    \[\begin{array}{*{20}{l}}
      {{D_f} = \left\{ {x \in \mathbb{R}:{x^2} – 4 \ne 0} \right\} = \mathbb{R}\backslash \left\{ { – 2,2} \right\}}&{\text{e}}&{{D_g} = \left\{ {x \in \mathbb{R}:x + 2 \ne 0} \right\} = \mathbb{R}\backslash \left\{ { – 2} \right\}}
    \end{array}\]
    As funções não são iguais, pois ${D_f} \ne {D_g}$.
0

Determine os números reais a, b e c

Operações com funções: Infinito 11 A - Parte 2 Pág. 200 Ex. 60

Enunciado

  1. Determine os números reais a, b e c tais que: \[\frac{3{{x}^{2}}-5x-7}{x-2}=ax+b+\frac{c}{x-2}\]
  2. Conjecture se o gráfico da função racional definida por \[f(x)=\frac{3{{x}^{2}}-5x-7}{x-2}\] tem uma assimptota oblíqua e, no caso afirmativo, indique a sua equação.

Resolução >> Resolução

  1.  Efectuando a divisão do polinómio $3{{x}^{2}}-5x-7$ por $x-2$ pela Regra de Ruffini, temos:
    \[\begin{array}{*{20}{c}}
      {}&3&{ – 5}&{ – 7} \\
      2&{}&6&2 \\
      {}&3&1&{ – 5}
    \end{array}\]
      Assim, temos: \[\frac{3{{x}^{2}}-5x-7}{x-2}=3x+1+\frac{-5}{x-2}\]
     Logo, $a=3\wedge b=1\wedge c=-5$.
f é outra função racional 0

f é outra função racional

Operações com funções: Infinito 11 A - Parte 2 Pág. 200 Ex. 59

Enunciado

f é uma função racional definida em $\mathbb{R}\backslash \left\{ -1,1 \right\}$ por \[f(x)=\frac{1}{1-{{x}^{2}}}\]

Encontre os reais a e b tais que, para todo o $x\ne 1\wedge x\ne -1$, \[f(x)=\frac{a}{1-x}+\frac{b}{1+x}\]

Resolução >> Resolução

Ora, \[\begin{array}{*{35}{l}}
   f(x) & = & \frac{a}{1-x}+\frac{b}{1+x}  \\
   {} & = & \frac{a(1+x)}{(1-x)(1+x)}+\frac{b(1-x)}{(1-x)(1+x)}  \\
   {} & = & \frac{(a-b)x+(a+b)}{1-{{x}^{2}}}  \\
\end{array}\]

Dado que \[f(x)=\frac{1}{1-{{x}^{2}}}\] vem:

\[\begin{array}{*{35}{l}}
   \left\{ \begin{array}{*{35}{l}}
   a-b=0  \\
   a+b=1  \\
\end{array} \right.…