Question

Não estou me ligando na relação da número (01).
Coloque um exemplo aí!

Answer 1

Boa noite Marcelo!

Solução!

$f(x)= \dfrac{x+1}{x-1}~~~~~x \in \mathbb{R}-\{0,1\}\\\\\\ Temos~~que~~provar~~que\\\\\\ f\left ( \frac{1}{x} \right )=-f(x)\\\\\\ Esse~~subconjunto~~acima~~e~~a~~restric\~ao,ou~~seja:\\\\ n\~ao~~se~~divide~~nada~~por~~zero.$



$\left ( \dfrac{1}{x} \right )=-f(x)\\\\\\ \left ( \dfrac{1}{0} \right )=-f(x) \Rightarrow \not \exists (N\~ao~~existe)$


$f\left ( \frac{1}{x} \right )=-f(x)\\\\\\ x= \dfrac{1}{x}\\\\\\Vamos~~agora~~substituir~~na~~fuc\~ao.\\\\\\ \dfrac{1+x}{ \dfrac{x}{~~\dfrac{1}{x} -1} }+1 = -\dfrac{x+1}{x-1}\\\\\\\ \dfrac{1+x}{ \dfrac{x}{ \dfrac{1-x}{x} } } = -\dfrac{x+1}{x-1}\\\\\\\ \frac{1+x}{x} . \frac{x}{1-x} = -\dfrac{x+1}{x-1}\\\\\\ \frac{1+x}{1-x}= -\dfrac{x+1}{x-1}\\\\\\ Organizando!\\\\\\\ \frac{x+1}{-x+1}= -\dfrac{x+1}{x-1}\\\\\\ Multiplicando~~o~~denominador~~por~~-1$


$\dfrac{x+1}{-1(x+1)}= - \dfrac{x+1}{x-1}\\\\\\ -1 \dfrac{x+1}{x-1}= - \dfrac{x+1}{x-1}\\\\\\ \boxed{Verdadeiro!~~f \left ( \dfrac{1}{x} \right )=-f(x)}$



Exercício 2

$f(x)= \dfrac{x+1}{x-1} \\\\\\\ g(x)=2x+3\\\\\\ D(fog)-\{-1\}\\\\\\ fog=f(g)x)\\\\\\ fog = \dfrac{(2x+3)+1}{(2x+3)-1}\\\\\\ fog = \dfrac{2x+3+1}{2x+3-1}\\\\\\ fog = \dfrac{2x+4}{2x+2}\\\\\\ Simplificando!\\\\\\\\ fog = \dfrac{(x+2)}{(x+1)}\\\\\\ fog = \dfrac{(x+2)}{(x+1)}$


$fog = \dfrac{(x+2)}{(x+1)}\\\\\ x+1 \neq 0\\\\\\ x \neq -1\\\\\\\\ Logo!\\\\\ Dominio~s\~ao~todos~os~reais~excluindo~~-1\\\\\ ~~pois~~n\~ao~existe~~divis\~ao~por~~0\\\\\\ Resposta!\\\\\ \boxed{D(fog)=\mathbb{R}-\{-1\}}$


Boa noite!
Bons estudos!