2≤|(|x|+1)/(x+3)|
To solve the inequality 2 ≤ |(|x| + 1)/(x + 3)|, we need to consider different cases based on the sign of the expression inside the absolute value.
Case 1: (x + 3) > 0
If (x + 3) > 0, we can remove the absolute value bars without changing the inequality. So we have:
2 ≤ (|x| + 1)/(x + 3)
To simplify the expression on the right-hand side, we can split it into two cases:
Subcase 1.1: x ≥ 0
If x ≥ 0, then |x| = x. So the inequality becomes:
2 ≤ (x + 1)/(x + 3)
To solve this inequality, we can multiply both sides by (x + 3) since it is positive:
2(x + 3) ≤ (x + 1)
Expanding and simplifying:
2x + 6 ≤ x + 1
x + 6 ≤ 1
x ≤ -5
However, this solution is not valid for this subcase because it contradicts our initial assumption that (x + 3) > 0. Therefore, there are no solutions for this subcase.
Subcase 1.2: x < 0
If x < 0, then |x| = -x. So the inequality becomes:
2 ≤ (-x + 1)/(x + 3)
To solve this inequality, we can multiply both sides by (x + 3) since it is positive:
2(x + 3) ≤ -x + 1
Expanding and simplifying:
2x + 6 ≤ -x + 1
3x ≤ -5
x ≤ -5/3
Therefore, the solutions for this subcase are -5/3 ≤ x < 0.
Case 2: (x + 3) < 0
If (x + 3) < 0, we have to keep the absolute value bars but change the inequality sign when removing them. So we have:
2 ≤ -(|x| + 1)/(x + 3)
To simplify the expression on the right-hand side, we can split it into two cases:
Subcase 2.1: x ≥ 0
If x ≥ 0, then |x| = x. So the inequality becomes:
2 ≤ -(x + 1)/(x + 3)
Multiplying both sides by (x + 3):
2(x + 3) ≤ -(x + 1)
Expanding and simplifying:
2x + 6 ≤ -x - 1
3x ≤ -7
x ≤ -7/3
Therefore, the solutions for this subcase are x ≤ -7/3.
Subcase 2.2: x < 0
If x < 0, then |x| = -x. So the inequality becomes:
2 ≤ -(-x + 1)/(x + 3)
Multiplying both sides by (x + 3):
2(x + 3) ≤ -(-x + 1)
Expanding and simplifying:
2x + 6 ≤ x - 1
x ≤ -7
Therefore, the solutions for this subcase are x ≤ -7.
In summary, combining the solutions from different cases and subcases, we have:
-5/3 ≤ x < 0 or x ≤ -7.
So these are the values of x that satisfy the given inequality.