what is the interval notation for solution set of the inequality x^2+16x is less than or equal to 0
x^2+16x < 0
x(x+16) < 0
You know the graph crosses the x-axis at x=0,-16. So,
x(x+16) < 0 if -16<x<0
To find the interval notation for the solution set of the inequality x^2 + 16x ≤ 0, we need to determine the values of x that satisfy the inequality.
First, let's find the critical points of the inequality by setting the expression x^2 + 16x equal to zero:
x^2 + 16x = 0
Factoring out an x, we get:
x(x + 16) = 0
So, either x = 0 or x + 16 = 0.
From this, we find two critical points: x = 0 and x = -16.
Next, we need to test the inequality in intervals between these critical points to determine which intervals satisfy the inequality.
We take a few test points from each interval and substitute them into the inequality to check if they make the inequality true or false.
Interval (-∞, -16):
Let's test x = -17:
(-17)^2 + 16(-17) = 289 - 272 = 17. Since 17 > 0, this interval does not satisfy the inequality.
Interval (-16, 0):
Let's test x = -10:
(-10)^2 + 16(-10) = 100 - 160 = -60. Since -60 < 0, this interval satisfies the inequality.
Interval (0, ∞):
Let's test x = 1:
1^2 + 16(1) = 1 + 16 = 17. Since 17 > 0, this interval does not satisfy the inequality.
Based on the test results, we can conclude that the solution set of the inequality x^2 + 16x ≤ 0 is the interval (-16, 0], which means that x is greater than -16 and less than or equal to 0.
Therefore, the interval notation for the solution set of the given inequality is (-16, 0].