Question

o conjunto solução da inequação
$( \frac{1}{2} ) {}^{x {}^{2} - 4 } \leqslant 8 {}^{x + 2}$
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Answer 1

Resposta:

$\sf \displaystyle \left ( \dfrac{1}{2} \right )^{x^{2} -4} \leq 8^{x +2} \quad \gets \text{\sf 8 para 2}}$

$\sf \displaystyle \left ( \dfrac{1}{2} \right )^{x^{2} -4} \leq \left( 2^3 \right)^{x +2} \quad \gets \text{\sf multiplicar os expoentes.}}$

$\sf \displaystyle \left ( \dfrac{1}{2} \right )^{x^{2} -4} \leq \left( 2 \right)^{3x +6} \quad \gets \text{\sf 2 para base 1/2 e inverte o sinal.}}$

$\sf \displaystyle \left ( \dfrac{1}{2} \right )^{x^{2} -4} \leq \left( \left ( \dfrac{1}{2} \right )^{-1} \right)^{3x +6} \quad \gets \text{\sf multiplicar os expoentes .}}$

$\sf \displaystyle \left ( \dfrac{1}{2} \right )^{x^{2} -4} \leq \left ( \dfrac{1}{2} \right )}^{-3x - 6} \quad \gets \text{\sf Cancelar a base 1/2 }}$

$\sf \displaystyle x^{2} - 4 \leq -3x - 6$

$\sf \displaystyle x^{2} + 3x - 4 + 6 \leq 0$

$\sf \displaystyle x^{2} + 3x + 2 \leq 0$

Determinar o Δ:

$\sf \displaystyle \Delta = b^2 -\:4ac$

$\sf \displaystyle \Delta = 3^2 -\:4 \cdot 1 \cdot 2$

$\sf \displaystyle \Delta = 9 - 8$

$\sf \displaystyle \Delta = 1$

Determinar as raízes da inequação:

$\sf \displaystyle x = \dfrac{-\,b \pm \sqrt{ \Delta } }{2a} = \dfrac{-\,3 \pm \sqrt{ 1 } }{2\cdot 1} = \dfrac{-\,3 \pm 1}{2} \Rightarrow\begin{cases} \sf x_1 = &\sf \dfrac{-\,3 + 1}{2} = \dfrac{2}{2} = -\:1 \\\\ \sf x_2 = &\sf \dfrac{-\,3 - 1}{2} = \dfrac{-\:4}{2} = - \:2\end{cases}$

$\boldsymbol{ \sf \displaystyle S=\{x\in\mathbb{R}\mid -\:2\leq x \leq -\;1 \}}$