You got 4 key x values: x = -1 (vertical asymptote), x = 0 (a solution), x = 3 (a solution), and x = 6 (vertical asymptote). Break everything down into intervals.
If x < -1, you have (-)(+) / (-)(-) = minus/plus = minus. Fail.
If -1 < x < 0, you have (-)(+) / (+)(-) = minus/minus = plus. Check.
If 0 < x < 3, you have (+)(+) / (+)(-) = plus/minus = minus. Fail.
If 3 < x < 6, you have (+)(+) / (+)(-) = plus/minus = minus. Fail.
If x > 6, you have (+)(+) / (+)(+) = plus/plus = plus. Check.
First thing to note: When you have a factor like (x - a), that factor is positive when x > a and negative when x < a. If you have a plus sign, that just means a is negative. For instance x + 2 > 0 when x > -2 and negative when x < -2.
OK, remembering that, let's factor.
Numerator:
x^3 - 6x^2 + 9x = x(x^2 - 6x + 9) = x(x - 3)^2. The x - 3 is squared, it's never negative. So the numerator is positive when x > 0 and negative when x < 0 (and 0 of course, when x = 0). It's also 0 when x - 3 is 0.
Denominator: x^2 - 5x - 6 = (x + 1)(x - 6). The critical points, where something changes sign, are at -1 from the first term and +6 from the second term.
When x < -1, both terms are negative. The product is > 0
When x is between -1 and 6 (more than -1, less than 6), the first term is positive, the second is negative. The product is < 0.
When x > 6, both terms are positive. The product is > 0.
Now let's put the numerator and denominator information together, breaking up the real numbers at the points -1, 0, and 6 where we find things change.
x < -1: Denominator > 0, numerator < 0. What is the sign of a negative divided by a negative?
-1 < x < 0: Denominator < 0, numerator < 0.
0 < x < 6: Denominator < 0, numerator > 0
x > 6: Denominator > 0, numerator > 0.
So in which of those combinations is the quotient > 0?
Some fine points to finish it off:
The inequality said ">=", so add the points where the numerator is 0.
You can't divide by 0, so make sure you omit the points x = -1 and x = 6 where the denominator is 0.
Answers & Comments
Verified answer
(x^3 - 6x^2 + 9x) / (x^2 - 5x - 6) >= 0
x(x - 3)^2 / (x + 1)(x - 6) >= 0
You got 4 key x values: x = -1 (vertical asymptote), x = 0 (a solution), x = 3 (a solution), and x = 6 (vertical asymptote). Break everything down into intervals.
If x < -1, you have (-)(+) / (-)(-) = minus/plus = minus. Fail.
If -1 < x < 0, you have (-)(+) / (+)(-) = minus/minus = plus. Check.
If 0 < x < 3, you have (+)(+) / (+)(-) = plus/minus = minus. Fail.
If 3 < x < 6, you have (+)(+) / (+)(-) = plus/minus = minus. Fail.
If x > 6, you have (+)(+) / (+)(+) = plus/plus = plus. Check.
Solution: -1 < x <= 0, x = 3, or x > 6.
(x³-6x²+9x) / (x²-5x-6) = 0
x(x-3)² / ((x+1)(x-6)) = 0
x(x-3)² = 0 and (x+1)(x-6) > 0
(x = 0 or x = 3) and (x < -1 or x > 6)
x > 6
or
x(x-3)² = 0 and (x+1)(x-6) < 0
(x = 0 or x = 3) and (-1 < x < 6)
-1 < x = 0 or x = 3
Solution:
-1 < x = 0 or x = 3 or x > 6
1. Factoring numerator and denominator
[x(x-3)²]/[(x+1)(x-6)]≥0
2. Existence domain
The inequality is NOT defined for x = -1 & x = 6, values that make the denominator null.
3. Put all the data in a grid of signs
.......-1.........0.........3........6......
----------------0+++++++++++++ (x)
++++++++++++++0++++++++ (x-3)²
-------X++++++++++++++++++ (x+1)
----------------------------------X+++ (x-6)
------X++++0-------0--------X+++ inequalitiy
We can state that inequality is satisfied in (-1,0] U (6,+oo)
or
-1<x≤0 V x>6
Find your zeroes
x^3 - 6x^2 + 9x =>
x * (x^2 - 6x + 9) =>
x * (x - 3)^2
x = 0 , 3
We have 3 domains to look through: (-inf , 0) , (0 , 3) and (3 , inf)
-10 * (-10 - 3)^2 = -10 * (-13)^2 = -10 * 169 = -1690
1 * (1 - 3)^2 = 1 * (-2)^2 = 1 * 4 = 4
5 * (5 - 3)^2 = 5 * 4 = 20
So, from -inf to 0, x^3 - 6x^2 + 9x is negative
From 0 to infinity, x^3 - 6x^2 + 9x is greater than or equal to 0
x^2 - 5x - 6 = 0
x = (5 +/- sqrt(25 + 24)) / 2
x = (5 +/- 7) / 2
x = 12/2 , -2/2
x = 6 , -1
Again, we have three domains: (-inf , -1) , (-1 , 6) , (6 , inf)
(-2)^2 - 5 * (-2) - 6 = 4 + 10 - 6 = 14 - 6 = 8
0^2 - 5 * 0 - 6 = -6
10^2 - 5 * 10 - 6 = 100 - 56 = 44
From -inf to -1 , x^2 - 5x - 6 is positive
From -1 to 6, x^2 - 5x - 6 is negative
From 6 to infinity, x^2 - 5x - 6 is positive.
When x = -1 and 6, x^2 - 5x - 6 is 0 and that means that we have values that will be undefined
So here are our domains that we have to look at:
(-inf , -1) x^3 - 6x^2 + 9x < 0 , x^2 - 5x - 6 > 0 , Negative
x = -1 , x^3 - 6x^2 + 9x < 0 , x^2 - 5x - 6 = 0 , Undefined
(-1 , 0) , x^3 - 6x^2 + 9x < 0 , x^2 - 5x - 6 < 0 , Positive
x = 0 , x^3 - 6x^2 + 9x = 0 , x^2 - 5x - 6 < 0 , 0
(0 , 3) , x^3 - 6x^2 + 9x > 0 , x^2 - 5x - 6 < 0 , Negative
x = 3 , x^3 - 6x^2 + 9x = 0 , x^2 - 5x - 6 < 0 , 0
(3 , 6) , x^3 - 6x^2 + 9x > 0 , x^2 - 5x - 6 < 0 , Negative
x = 6 , x^3 - 6x^2 + 9x > 0 , x^2 - 5x - 6 = 0 , Undefined
(6 , infinity) , x^3 - 6x^2 + 9x > 0 , x^2 - 5x - 6 > 0 , Positive
(-1 , 0]U[3]U(6 , infinity)
Confirmed with Wolfram Alpha
https://www.wolframalpha.com/input/?i=(x%C2%B3-6x%...
First thing to note: When you have a factor like (x - a), that factor is positive when x > a and negative when x < a. If you have a plus sign, that just means a is negative. For instance x + 2 > 0 when x > -2 and negative when x < -2.
OK, remembering that, let's factor.
Numerator:
x^3 - 6x^2 + 9x = x(x^2 - 6x + 9) = x(x - 3)^2. The x - 3 is squared, it's never negative. So the numerator is positive when x > 0 and negative when x < 0 (and 0 of course, when x = 0). It's also 0 when x - 3 is 0.
Denominator: x^2 - 5x - 6 = (x + 1)(x - 6). The critical points, where something changes sign, are at -1 from the first term and +6 from the second term.
When x < -1, both terms are negative. The product is > 0
When x is between -1 and 6 (more than -1, less than 6), the first term is positive, the second is negative. The product is < 0.
When x > 6, both terms are positive. The product is > 0.
Now let's put the numerator and denominator information together, breaking up the real numbers at the points -1, 0, and 6 where we find things change.
x < -1: Denominator > 0, numerator < 0. What is the sign of a negative divided by a negative?
-1 < x < 0: Denominator < 0, numerator < 0.
0 < x < 6: Denominator < 0, numerator > 0
x > 6: Denominator > 0, numerator > 0.
So in which of those combinations is the quotient > 0?
Some fine points to finish it off:
The inequality said ">=", so add the points where the numerator is 0.
You can't divide by 0, so make sure you omit the points x = -1 and x = 6 where the denominator is 0.
(x³ - 6x² + 9x) / (x² - 5x - 6) ≥ 0
(x³ - 6x² + 9x) ≥ 0
(x - 3)^2 x ≥ 0
Solution:
x ≥ 0
easy as pie