(x+1) (x-2) (x-4) <0
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21 января 2025 15:40
To solve the inequality ((x+1)(x-2)(x-4) < 0), follow these steps:
Find the roots: Set the expression equal to zero to find the critical points: [ (x+1)(x-2)(x-4) = 0 ] The roots are (x = -1), (x = 2), and (x = 4).
Determine the intervals: The critical points divide the number line into the following intervals:
- ((-\infty, -1))
- ((-1, 2))
- ((2, 4))
- ((4, \infty))
Test each interval: Choose a test point from each interval to determine the sign of the product in that interval.
Interval ((-\infty, -1)) (test point (x = -2)): [ (-2 + 1)(-2 - 2)(-2 - 4) = (-1)(-4)(-6) = -24 < 0 ] (This interval satisfies the inequality.)
Interval ((-1, 2)) (test point (x = 0)): [ (0 + 1)(0 - 2)(0 - 4) = (1)(-2)(-4) = 8 > 0 ] (This interval does not satisfy the inequality.)
Interval ((2, 4)) (test point (x = 3)): [ (3 + 1)(3 - 2)(3 - 4) = (4)(1)(-1) = -4 < 0 ] (This interval satisfies the inequality.)
Interval ((4, \infty)) (test point (x = 5)): [ (5 + 1)(5 - 2)(5 - 4) = (6)(3)(1) = 18 > 0 ] (This interval does not satisfy the inequality.)
Consider the critical points: The expression is zero at (x = -1), (x = 2), and (x = 4), which means it does not satisfy the strict inequality (< 0) at these points.
Combine the results: The intervals where the product is negative are: [ (-\infty, -1) \quad \text{and} \quad (2, 4) ]
Thus, the solution for the inequality ((x+1)(x-2)(x-4) < 0) is: [ (-\infty, -1) \cup (2, 4) ]
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