Question

A função q : R → R é tal que, para qualquer valor real de x, vale q(3 – 2x) = x + 3.
Calcule o valor de q(3) + q(4) – q(5).

Answer 1

Resposta:

$\boxed{\mathtt{q(3) + q(4) - q(5) = 7/2}}$

Explicação passo-a-passo:

$\\ \displaystyle \mathsf{q(3 - 2x) = x + 3} \\\\ \mathsf{q(3 - 2x) = x + 3 + (2x - 2x)} \\\\ \mathsf{q(3 - 2x) = (x + 2x) + (3 - 2x)} \\\\ \mathsf{q(3 - 2x) = (3 - 2x) + 3x}$

Por conseguinte, considere $\displaystyle \mathtt{3 - 2x = \lambda}$. Com efeito, $\displaystyle \mathtt{x = \frac{3 - \lambda}{2}}$. Daí,

$\\ \displaystyle \mathsf{q(3 - 2x) = (3 - 2x) + 3x} \\\\ \mathsf{q(\lambda) = \lambda + 3 \cdot \frac{(3 - \lambda)}{2}} \\\\ \mathsf{q(\lambda) = \frac{2\lambda + 9 - 3\lambda}{2}} \\\\ \boxed{\mathsf{q(\lambda) = \frac{9 - \lambda}{2}}}$

Por fim, temos:

$\\ \displaystyle \mathsf{q(\lambda) = \frac{9 - \lambda}{2} \Rightarrow \begin{cases} \bullet \quad \mathsf{q(3) = \frac{9 - 3}{2} \Rightarrow \boxed{\mathsf{q(3) = 3}}} \\\\ \bullet \quad \mathsf{q(4) = \frac{9 - 4}{2} \Rightarrow \boxed{\mathsf{q(4) = \frac{5}{2}}}} \\\\ \bullet \quad \mathsf{q(5) = \frac{9 - 5}{2} \Rightarrow \boxed{\mathsf{q(5) = 2}}} \end{cases}}$

Logo,

$\\ \displaystyle \mathsf{q(3) + q(4) - q(5) =} \\\\ \mathsf{3 + \frac{5}{2} - 2 =} \\\\ \mathsf{1 + \frac{5}{2} =} \\\\ \boxed{\boxed{\mathsf{\frac{7}{2}}}}$