Se g(x) = 1-x e (fog)(x) = 
, com x≠0, então f(
) vale:
a) 1
b) 
c) 4
d) -
e) -4
Soluções para a tarefa
Resposta:
fog(x)=
x
1−x
f(g(x))=
x
1−x
f(1−x)=
x
1−x
Vamos chamar 1 - x de k, e isolar x:
\begin{lgathered}1-x=k\\x=1-k\end{lgathered}
1−x=k
x=1−k
Logo:
\begin{lgathered}f(1-x)=\dfrac{1-x}{x}\\\\\\f(k)=\dfrac{1-(1-k)}{1-k}\\\\\\f(k)=\dfrac{1-1+k}{1-k}\\\\\\f(k)=\dfrac{k}{1-k}\end{lgathered}
f(1−x)=
x
1−x
f(k)=
1−k
1−(1−k)
f(k)=
1−k
1−1+k
f(k)=
1−k
k
Trocando k por x:
\boxed{\boxed{f(x)=\dfrac{x}{1-x}}}
f(x)=
1−x
x
________________________
\begin{lgathered}f(x)=\dfrac{x}{1-x}\\\\\\f\left(\dfrac{4}{3}\right)=\dfrac{\left(\frac{4}{3}\right)}{1-\left(\frac{4}{3}\right)}\\\\\\f\left(\dfrac{4}{3}\right)=\dfrac{\left(\frac{4}{3}\right)}{\left(-\frac{1}{3}\right)}\\\\\\f\left(\dfrac{4}{3}\right)=-\dfrac{\left(\frac{4}{3}\right)}{\left(\frac{1}{3}\right)}\\\\\\f\left(\dfrac{4}{3}\right)=-\dfrac{4}{3}*\dfrac{3}{1}\\\\\\\boxed{\boxed{f\left(\dfrac{4}{3}\right)=-4}}\end{lgathered}
f(x)=
1−x
x
f(
3
4
)=
1−(
3
4
)
(
3
4
)
f(
3
4
)=
(−
3
1
)
(
3
4
)
f(
3
4
)=−
(
3
1
)
(
3
4
)
f(
3
4
)=−
3
4
∗
1
3
f(
3
4
)=−4
______________________________________
Se não quiser achar f(x) primeiro, podemos achar o valor de x que faça com que 1 - x seja igual a 4/3
\begin{lgathered}f(1-x)=f\left(\dfrac{4}{3}\right)\\\\\\1-x=\dfrac{4}{3}\\\\\\x=1-\dfrac{4}{3}\\\\\\\boxed{\boxed{x=-\dfrac{1}{3}}}\end{lgathered}
f(1−x)=f(
3
4
)
1−x=
3
4
x=1−
3
4
x=−
3
1
Então:
\begin{lgathered}f(1-x)=\dfrac{1-x}{x}\\\\\\f\left(1-\left[-\dfrac{1}{3}\right]\right)=\dfrac{1-\left(-\frac{1}{3}\right)}{\left(-\frac{1}{3}\right)}~~~~\therefore~~~~\boxed{\boxed{f\left(\dfrac{4}{3}\right)=-4}}\end{lgathered}
f(1−x)=
x
1−x
f(1−[−
3
1
])=
(−
3
1
)
1−(−
3
1
)
∴
f(
3
4
)=−4