How can you solve this inequality using an algebraic method.
-x^3+2x^2+4x-8 > or equal to 0
(x-2)(-x+2)(x+2) > or equal to 0
you have correctly determined that
y = -(x-2)^2 (x+2)
You know the graph crosses the axis at x = -2 and that it just touches the x-axis at x=2 (because of the double root)
Since for large negative values of x, all 3 factors are negative,
y > 0 for x < -2
So, y < 0 for -2 < x < 2 because only one of the factors is negative
But y is also < 0 for x > 2 because all three factors are positive
So, y >= 0 for x <= -2
A nice graph can be seen at
http://www.wolframalpha.com/input/?i=-x^3%2B2x^2%2B4x-8
To solve the inequality -x^3 + 2x^2 + 4x - 8 ≥ 0 using an algebraic method, we can use the concept of sign charts.
1. First, find the critical points of the inequality. These are the values of x where the expression is equal to zero. In this case, the critical points are x = -2, x = 2, and x = -√2 (approximately -1.414).
2. Next, choose test values for each interval determined by the critical points. You can choose simple values, such as -3, -1, 0, 1, and 3, to test.
3. Evaluate the expression (-x+2)(x-2)(x+2) for each test value and determine the sign.
- For x < -2, choose -3. Substitute -3 into the expression: (-(-3)+2)(-3-2)(-3+2) = (5)(-5)(-1) = 25. The expression is positive.
- For -2 < x < -√2, choose -1. Substitute -1 into the expression: (-(1)+2)(-1-2)(-1+2) = (1)(-3)(1) = -3. The expression is negative.
- For -√2 < x < 2, choose 0. Substitute 0 into the expression: (-(0)+2)(0-2)(0+2) = (2)(-2)(2) = -8. The expression is negative.
- For 2 < x, choose 3. Substitute 3 into the expression: (-(3)+2)(3-2)(3+2) = (-1)(1)(5) = -5. The expression is negative.
4. Represent the sign information using a sign chart:
(-∞, -2) | (-2, -√2) | (-√2, 2) | (2, ∞)
+ | - | - | -
5. Finally, determine the solution by considering the sign chart. The expression (-x+2)(x-2)(x+2) is greater than or equal to zero when the signs are positive or zero. Therefore, the solution is when x belongs to the intervals (-∞, -2] and [2, ∞).
In interval notation, the solution is (-∞, -2] ∪ [2, ∞).