Rize S. answered • 03/23/23
MISM + 25 Yrs Exp: Algebra 1 Pro
To solve the inequality |3x^3+6x^2-7x-14| ≥ |3-x|, we need to consider two cases:
Case 1: (3-x) ≥ 0, or x ≤ 3
In this case, we can simplify the inequality by removing the absolute value symbols:
3x^3+6x^2-7x-14 ≥ 3-x
Simplifying this expression, we get:
3x^3+6x^2-6x-11 ≥ 0
We can solve this inequality by factoring the left-hand side:
(3x^2+9x+11)(x-1) ≥ 0
The quadratic factor 3x^2+9x+11 has no real roots, since its discriminant is negative. Therefore, the sign of the expression is the same as the sign of the linear factor (x-1).
Since x ≤ 3 in this case, we have x-1 ≤ 2. Therefore, the inequality is satisfied for all x ≤ 3.
Case 2: (3-x) < 0, or x > 3
In this case, we can simplify the inequality by changing the sign of the right-hand side and removing the absolute value symbols:
3x^3+6x^2-7x-14 ≥ x-3
Simplifying this expression, we get:
3x^3+6x^2-6x-11 ≥ 0
This is the same cubic expression as in Case 1. Therefore, the inequality is satisfied for all x > 3 as well.
Putting these two cases together, we get the solution:
x ≤ 3 or x > 3
In interval notation, this can be written as:
(-∞, 3] ∪ (3, ∞)
