Dadas as funções $f$ e $g$
Mais funções: Aleph 11 - Volume 2 Pág. 137 Ex. 7
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:
- $\left( {f \circ f} \right)\left( 1 \right)$
- $\left( {g \circ g} \right)\left( 2 \right)$
- $\left( {f \circ g} \right)\left( 2 \right)$
- $\left( {g \circ f} \right)\left( 2 \right)$
\[\begin{array}{*{20}{c}} {f\left( x \right) = 2x + 3}&{\text{e}}&{g\left( x \right) = – {x^2} + 5} \end{array}\]
- $\left( {f \circ f} \right)\left( 1 \right) = f\left( {f\left( 1 \right)} \right) = f\left( {2 \times 1 + 3} \right) = f\left( 5 \right) = 2 \times 5 + 3 = 13$
- $\left( {g \circ g} \right)\left( 2 \right) = g\left( {g\left( 2 \right)} \right) = g\left( { – {2^2} + 5} \right) = g\left( 1 \right) = – {1^2} + 5 = 4$
- $\left( {f \circ g} \right)\left( 2 \right) = f\left( {g\left( 2 \right)} \right) = f\left( { – {2^2} + 5} \right) = f\left( 1 \right) = 2 \times 1 + 3 = 5$
- $\left( {g \circ f} \right)\left( 2 \right) = g\left( {f\left( 2 \right)} \right) = g\left( {2 \times 2 + 3} \right) = g\left( 7 \right) = – {7^2} + 5 = – 44$





