x⁴ - 11x³ + 18x² ≥ 0
Factor out x²
x²(x² - 11x + 18) ≥ 0
Factor the trinomial in the parentheses:
x²(x - 2)(x - 9) ≥ 0
The critical numbers are 0, 2 and 9.
They are also solution because when
substituted for x the cause the left
side to equal 0.
So we put those on a number line with
a darkened circle:
----------☻-----☻--------------------☻---------
-3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12
We choose easiest test values in between those
darkened circles.
for the interval x < 0 we choose -1 and substitute
it in the inequality:
x²(x - 2)(x - 9) ≥ 0
(-1)²(-1 - 2)(-1 - 9) ≥ 0
1(-3)(-10) ≥ 0
30 ≥ 0
That's true so we shade the interval x < 0
<=========☻-----☻--------------------☻---------
-3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12
for the interval 0 < x < 2 we choose test value 1
and substitute it in the inequality:
x²(x - 2)(x - 9) ≥ 0
(1)²(1 - 2)(1 - 9) ≥ 0
1(-1)(-8) ≥ 0
30 ≥ 0
That's true so we shade the interval x < 0
<=========☻=====☻--------------------☻---------
-3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12
We might as well delete the darkened circle at 0
since now we know it is shaded anyway:
<===============☻--------------------☻---------
-3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12
for the interval 2 < x < 9 we choose test value 3
and substitute it in the inequality:
x²(x - 2)(x - 9) ≥ 0
(3)²(3 - 2)(3 - 9) ≥ 0
9(1)(-6) ≥ 0
-54 ≥ 0
That's false so we do not shade the interval 2 < x < 9.
So we still have:
<===============☻--------------------☻---------
-3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12
for the interval x > 9 we choose 10 and substitute
it in the inequality:
x²(x - 2)(x - 9) ≥ 0
(10)²(10 - 2)(10 - 9) ≥ 0
100(8)(1) ≥ 0
100 ≥ 0
That's true so we shade the interval x > 9.
<===============☻--------------------☻========>
-3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12
That's the number line graph. The interval
notation for that is:
Edwin