Question 1093348: could you please help me with this equation in interval notation
(x-9)(x+3)(x+6)>(or equal) 0
I tried and got [9,infinity) but I don't know if I am supposed to use the sign diagram where it would be positive inbetween (-6,-3) U (9,infinity)
Found 4 solutions by richwmiller, josgarithmetic, ikleyn, greenestamps:
Answer by richwmiller(17219)   (Show Source):
Answer by josgarithmetic(39616)  (Show Source):
You can put this solution on YOUR website! (x-9)(x+3)(x+6)>(or equal) 0
(x-9)(x+3)(x+6)>=0
Three critical x-values are -6, -3, and 9.
The intervals on the real number line you want to check are (-infinity, -6], [-6, -3], [-3, 9], and [9, infinity).
Check any value within each interval to find if that interval works or not works.
(-infinity, -6] [-6, -3] [-3, 9] [9, infinity)
signs (-)(-)(-) (-)(-)(+) (-)(+)(+) (+)(+)(+)
result sign (-) (+) (-) (+)
FALSE TRUE FALSE TRUE
The solution set, of all the numbers that can make the equation true is
[-6,-3] U [9, infinity).
NOTE: the form of the bracket for the interval notation, is important. Square bracket INCLUDES the boundary value. Curved bracket EXCLUDES the boundary value.
Answer by ikleyn(52775)  (Show Source):
Answer by greenestamps(13198)  (Show Source):
You can put this solution on YOUR website!
While the process that is normally taught for finding where the function value is positive or negative is to try a test point in each interval (as shown in one of the earlier responses), there is a much faster and easier way to do it. Just start on one end of the number line and determine the value of the function where you start. Then walk along the number line and see what happens each time you pass one of the critical points.
I usually start with a large positive value for x (on the right end of the number line) and walk left; but it might be easier to understand if we walk left to right.
So start with a "large negative" value for x. Clearly all three factors will be negative, so the function value will be negative. Then as we walk along the number line to the right, the sign of the function changes at every critical point, because at each of those points a single factor changes sign.
Our critical points are where the three factors are equal to 0 -- at -6, -3, and 9. So we start with the function negative for "large negative" values of x. The sign of the function then changes to positive when we pass x=-6, changes back to negative when we pass x=-3, and finally changes to positive when we pass x=9.
Note that with the "greater than or equal to" in the inequality, the endpoints of the intervals in the solution set will be included. So the solution set is
[-6,-3] U [9,infinity)
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