How do you solve and write the following in interval notation: #(x-2)/(x+1)

Question

How do you solve and write the following in interval notation:  #(x-2)/(x+1) <=0#?

Avery Alexander · Accepted Answer

The solution is #x in (-1,2]#

Let #f(x)=(x-2)/(x+1)#

We build a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaaa)##-1##color(white)(aaaaaa)##2##color(white)(aaaaaaa)##+oo#

#color(white)(aaaa)##x+1##color(white)(aaaaa)##-##color(white)(aaaa)##||##color(white)(aa)##+##color(white)(aa)##0##color(white)(aaa)##+#

#color(white)(aaaa)##x-2##color(white)(aaaaa)##-##color(white)(aaaa)##||##color(white)(aa)##-##color(white)(aa)##0##color(white)(aaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##+##color(white)(aaaa)##||##color(white)(aa)##-##color(white)(aa)##0##color(white)(aaa)##+#

Therefore,

#f(x)<=0# when #x in (-1,2]#

graph{(x-2)/(x+1) [-14.24, 14.24, -7.12, 7.12]}

Abigail Allen · Answer

undefined To solve and write the inequality $ \frac{x - 2}{x + 1} \leq 0 $ in interval notation, follow these steps: 1. Find the critical points by setting the numerator and denominator equal to zero and solving for $ x $. 2. Determine the sign of the expression $ \frac{x - 2}{x + 1} $ in each interval created by the critical points. 3. Express the solution in interval notation based on the signs of the expression. The critical points are $ x = 2 $ and $ x = -1 $. Test each interval: - When $ x < -1 $, $ \frac{x - 2}{x + 1} > 0 $ - When $ -1 < x < 2 $, $ \frac{x - 2}{x + 1} < 0 $ - When $ x > 2 $, $ \frac{x - 2}{x + 1} > 0 $ So, the solution in interval notation is: $ (-\infty, -1) \cup (2, \infty) $.

How do you solve and write the following in interval notation: #(x-2)/(x+1) &lt;=0#?

How do you solve and write the following in interval notation: #(x-2)/(x+1) <=0#?