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2.2.1: Ejercicios 2.2
https://espanol.libretexts.org/Matematicas/Precalculo_y_Trigonometria/Prec%C3%A1lculo_APEX_(Chapman%2C_Herald_y_Libertini)/02%3A_Habilidades_b%C3%A1sicas_para_el_c%C3%A1lculo/2.02%3A_Resolver_desigualdades/2.2.01%3A_Ejercicios_2.2
No, la declaración $x^2 >0$ is always true, but has a break point at $x=0$ . $x \geq -5$ and $x>2$ $x \geq -5$ and $x<2$ $x \leq -5$ and $x>2$ $x \geq -5$ or $x>2$ $3 \leq x <4$ or...No, la declaración $x^2 >0$ is always true, but has a break point at $x=0$ . $x \geq -5$ and $x>2$ $x \geq -5$ and $x<2$ $x \leq -5$ and $x>2$ $x \geq -5$ or $x>2$ $3 \leq x <4$ or $x>4$ $5 < x \leq 6$ or $7 \leq x <8$ $\displaystyle x \in (-\infty,-5) \cup (3,\infty)$ $\displaystyle x \in \Big[\frac{7-\sqrt{89}}{2}, \frac{7+\sqrt{89}}{2} \Big]$ $\displaystyle x \in (-\infty,2) \cup [7, \infty)$ $\displaystyle x \in (-\infty,-4)\cup [-2,-1] \cup (4,\infty)$

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