Resposta:
[tex]\sf \displaystyle \left ( \dfrac{1}{2} \right )^{x^{2} -4} \leq 8^{x +2} \quad \gets \text{\sf 8 para 2}}[/tex]
[tex]\sf \displaystyle \left ( \dfrac{1}{2} \right )^{x^{2} -4} \leq \left( 2^3 \right)^{x +2} \quad \gets \text{\sf multiplicar os expoentes.}}[/tex]
[tex]\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.}}[/tex]
[tex]\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 .}}[/tex]
[tex]\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 }}[/tex]
[tex]\sf \displaystyle x^{2} - 4 \leq -3x - 6[/tex]
[tex]\sf \displaystyle x^{2} + 3x - 4 + 6 \leq 0[/tex]
[tex]\sf \displaystyle x^{2} + 3x + 2 \leq 0[/tex]
Determinar o Δ:
[tex]\sf \displaystyle \Delta = b^2 -\:4ac[/tex]
[tex]\sf \displaystyle \Delta = 3^2 -\:4 \cdot 1 \cdot 2[/tex]
[tex]\sf \displaystyle \Delta = 9 - 8[/tex]
[tex]\sf \displaystyle \Delta = 1[/tex]
Determinar as raízes da inequação:
[tex]\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}[/tex]
[tex]\boldsymbol{ \sf \displaystyle S=\{x\in\mathbb{R}\mid -\:2\leq x \leq -\;1 \}}[/tex]
