Answer:
x ∈ (-2, -1) ∪ (1/3, ∞)
Step-by-step explanation:
To solve this problem we must factor the expression that is shown in the Numerator of the inequality.
So, we have:
[tex]3x^2 +2x -1 = 0\\3(x ^ 2+\frac{2}{3}x -\frac{1}{3}) = 0[/tex]
[tex]3(x ^ 2+\frac{2}{3}x -\frac{1}{3}) = 0\\[/tex]
We should look for two numbers that add 2/3 as a result and multiply as a result -1/3
These numbers are -1/3 and 1
Then:
[tex]3(x ^ 2+\frac{2}{3}x -\frac{1}{3}) = 3(x-\frac{1}{3})(x+1)[/tex]
So the roots are:
[tex]x = -\frac{1}{3}\\\\x = 1[/tex]
Now the expression is as follows:
[tex]\frac{3(x-\frac{1}{3})(x+1)}{(x + 2)}>0[/tex]
Now we use the study of signs to solve this inequality.
We have 3 roots for the polynomials that compose the expression:
[tex]x = -\frac{1}{3}\\\\x = 1\\\\x=-2[/tex]
We know that x = -2 is not in the domain of the function because it makes the denominator equal to zero
With these roots we make the study of signs:
Observe the attached image.
Note that:
[tex](x-\frac{1}{3})>0[/tex] when [tex]x>\frac{1}{3}[/tex]
[tex](x + 2)>0[/tex] when [tex]x>-2[/tex]
[tex](x + 1)>0[/tex] when [tex]x>-1[/tex]
Finally after the study of signs we can reach the conclusion that:
x ∈ (-2, -1) ∪ (1/3, ∞)
