which shows all the critical points for the inequality $\frac{x^{2}-4}{x^{2}-5x + 6}<0$?\n$x=-2$ and $x =…

which shows all the critical points for the inequality $\frac{x^{2}-4}{x^{2}-5x + 6}<0$?\n$x=-2$ and $x = 2$\n$x = 2$ and $x = 3$\n$x=-3,x=-2$, and $x = 2$\n$x=-2,x = 2$, and $x = 3$

which shows all the critical points for the inequality $\frac{x^{2}-4}{x^{2}-5x + 6}<0$?\n$x=-2$ and $x = 2$\n$x = 2$ and $x = 3$\n$x=-3,x=-2$, and $x = 2$\n$x=-2,x = 2$, and $x = 3$

Answer

Explanation:

Step1: Factor the numerator and denominator

The numerator $x^{2}-4=(x + 2)(x - 2)$. The denominator $x^{2}-5x + 6=(x-2)(x - 3)$. So the inequality becomes $\frac{(x + 2)(x - 2)}{(x-2)(x - 3)}<0$.

Step2: Find the values that make the numerator and denominator zero

Set the numerator $(x + 2)(x - 2)=0$, we get $x=-2$ or $x = 2$. Set the denominator $(x-2)(x - 3)=0$, we get $x=2$ or $x=3$. But we need to consider the domain - values that make the denominator zero are not in the domain of the rational - function. After canceling out the common factor $(x - 2)$ (for $x\neq2$), the function is $y=\frac{x + 2}{x - 3}$, and the critical points are the values that make the numerator zero ($x=-2$) and the values that make the original denominator zero ($x = 3$), and we also need to consider the value from the original non - simplified numerator zero ($x = 2$).

Answer:

$x=-2,x = 2$, and $x=3$