How do I construct a cubic inequality with specific solution sets?

[iluvrotinsaag]

Math problems!

I understand how to solve this inequality, but i do not know what sign to put (greater than or less than) in my answer. I would really appreciate it if someone could explain this concept to me. THanks

Write a cubic inequality that has this solution:
-2<X<0, x>3
(X)(X+2)(X-3)
(X^2 +2X) (X-3)
X^3 - X^2 - 6X (< or >) 0

 

[cookiemonster]

Whoa, hold on. What's the inequality?

x>3 into the expression x^3 - x^2 - 6x?

What's the -2<x<0?

cookiemonster

 

[iluvrotinsaag]

teh cubic inequality is:
-2<X<0 and x>3

 

[cookiemonster]

I've never heard of a cubic inequality before (other than a cubic polynomial used in an inequality), so I'll just butt out of this before I make a fool of myself again.

cookiemonster

 

[Damned charming :)]

you can always check if you have the inequality the right way round you do not need a method for checking if its "<" or ">" other than subbing in points

for example
when x=4, the statement x> 3 is true
and the statement 4^3-4^2-6*4=24 > 0 is true so the answer better be
x^3 - x^2 - 6x> 0
not x^3 - x^2 - 6x <0.


There is no mystery here just sketch the curve y=x^3 - x^2 - 6x
and notice there are two regions above the x-axis, one region has the property that all x values are between -2 and 0 (-2< x< 0) and the other region is x>3

 

[HallsofIvy]

Apparently the problem is not to "solve an inequality" but to construct an inequality having all numbers between -2 and 0 and all numbers above 3 as solutions.

Start by constructing an equation:

(x-(-2))(x-0)(x-3)= (x+2)(x)(x-3)= x³+ 5x²+ 6x= 0 is 0 at exactly -2, 0, and 3. Since a polynomial is continuous, it can change from positive to negative and vice-versa only where it is 0.

Now check what happens in between: at x= 1, for example,
1³+ 5(1²)+ 6(1)= 1+ 5+ 6= 11> 0. Therefore, the inequality x³+ 5x²+ 6x> 0 has x<-2 and 0< x< 3 as solution set.

IF the problem had asked for the complement set: -2< x< 0, x> 3, then we would have tried x= -1, say, and found that
(-1)^{3+ 5(-1)2+ 6(-1)= -1+5-6= -2< 0.
The inequality x3+ 5x2+ 6x< 0 has that solution set.}