Respuesta :
Answer:
Critical points are: 2 , -2 and 3.
Step-by-step explanation:
We have been given the expression:
[tex]\frac{x^2-4}{x^2-5x+6}<0[/tex]
We will factorize the given expression:
Using [tex]a^2-b^2=(a+b)(a-b)[/tex]
[tex]\frac{(x+2)(x-2)}{x^2-3x-2x+6}<0[/tex]
[tex]\Rightarrow \frac{(x+2)(x-2)}{x(x-3)-2(x-3)}<0[/tex]
[tex]\Rightarrow \frac{(x+2)(x-2)}{(x-2)(x+3)}<0[/tex]
Critical point are those values of an expression when it is equal to zero
Hence, the critical points are: 2,-2 and -3.
Answer:
d. x = –2, x = 2, and x = 3 are critical points.
Step-by-step explanation:
Given : [tex]\frac{x^{2}-4 }{x^{2}-5x+6}[/tex] < 0.
To find :What are the critical points for the inequality.
Solution : We have given that
[tex]\frac{x^{2}-4 }{x^{2}-5x+6}[/tex] < 0.
Using [tex]a^{2}-b^{2} = (a+b)(a-b)[/tex]
[tex]\frac{(x-2)(x+2) }{x^{2}-5x+6}[/tex] < 0.
Now, on factoring denominator
[tex]x^{2} -5x+6[/tex]
[tex]x^{2} -3x -2x+6[/tex]
Taking commom
x( x- 3) -2(x -3)
On grouping (x-3)(x -2)
[tex]\frac{(x-2)(x+2) }{(x-3)(x-2)}<0[/tex] .
We need to find critical point ,
Critical point on which expression is zero
Then x = 2, -2, 3 are critical point.
Therefore, d. x = –2, x = 2, and x = 3 are critical points.
