Question

Dadas as funções reais f(x) e g(x) definidas por: f(x)=x²-x-20 g(x) = 1 - 2x. Indique a alternativa que corresponde aos resultados de g(f(-2)) e f(g(-2))​

Answer 1

✅ Após resolver os cálculos, concluímos que os valores numéricos das referidas composição de funções são, respetivamente:

\Large\displaystyle\text{$\begin{gathered}\boxed{\boxed{\:\:\:\bf g(f(-2)) = 29\:\:\:e\:\:\:f(g(-2)) = 0\:\:\:}}\end{gathered}$}

g(f(−2))=29ef(g(−2))=0

Sejam as funções polinomiais:

\begin{gathered}\Large\begin{cases} f(x) = x^{2} - x - 20\\g(x) = 1 - 2x\\ g(f(-2)) = \:?\\f(g(-2)) = \:?\end{cases}\end{gathered}

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f(x)=x

2

−x−20

g(x)=1−2x

g(f(−2))=?

f(g(−2))=?

Organizando as equações, temos:

\begin{gathered}\Large\begin{cases} f(x) = x^{2} - x - 20\\g(x) = -2x + 1\\g(f(-2)) = \:?\\f(g(-2)) = \:?\end{cases}\end{gathered}

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f(x)=x

2

−x−20

g(x)=−2x+1

g(f(−2))=?

f(g(−2))=?

Então, temos:

Calculando g(f(-2)):

\Large\displaystyle\text{$\begin{gathered} g(f(-2)) = -2[f(-2)] + 1\end{gathered}$}

g(f(−2))=−2[f(−2)]+1

\Large\displaystyle\text{$\begin{gathered} = -2[(-2)^{2} - (-2) - 20] + 1\end{gathered}$}

=−2[(−2)

2

−(−2)−20]+1

\Large\displaystyle\text{$\begin{gathered} = -2[4 + 2 - 20] + 1\end{gathered}$}

=−2[4+2−20]+1

\Large\displaystyle\text{$\begin{gathered} = -2[-14] + 1\end{gathered}$}

=−2[−14]+1

\Large\displaystyle\text{$\begin{gathered} = 28 + 1\end{gathered}$}

=28+1

\Large\displaystyle\text{$\begin{gathered} = 29\end{gathered}$}

=29

\Large\displaystyle\text{$\begin{gathered} \therefore\:\:\:g(f(-2)) = 29\end{gathered}$}

∴g(f(−2))=29

Calculando f(g(-2)):

\Large\displaystyle\text{$\begin{gathered} f(g(-2)) = (g(-2))^{2} - (g(-2)) - 20\end{gathered}$}

f(g(−2))=(g(−2))

2

−(g(−2))−20

\large\displaystyle\text{$\begin{gathered} = (-2\cdot(-2) + 1)^{2} - (-2\cdot(-2) + 1) - 20\end{gathered}$}

=(−2⋅(−2)+1)

2

−(−2⋅(−2)+1)−20

\Large\displaystyle\text{$\begin{gathered} = (4 + 1)^{2} - (4 + 1) - 20\end{gathered}$}

=(4+1)

2

−(4+1)−20

\Large\displaystyle\text{$\begin{gathered} = 5^{2} - 5 - 20\end{gathered}$}

=5

2

−5−20

\Large\displaystyle\text{$\begin{gathered} = 25 - 5 - 20\end{gathered}$}

=25−5−20

\Large\displaystyle\text{$\begin{gathered} = 0\end{gathered}$}

=0

\Large\displaystyle\text{$\begin{gathered} \therefore\:\:\:f(g(-2)) = 0\end{gathered}$}

∴f(g(−2))=0

\LARGE\displaystyle\text{$\begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Bons \:estudos!!\:\:\:Boa\: sorte!!\:\:\:}}}\end{gathered}$}

Bonsestudos!!Boasorte!!