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Вопрос от Анонимного юзера 21 января 2025 15:40

(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:

  1. 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).

  2. Determine the intervals: The critical points divide the number line into the following intervals:

    • ((-\infty, -1))
    • ((-1, 2))
    • ((2, 4))
    • ((4, \infty))
  3. 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.)

  4. 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.

  5. 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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(x+1) (x-2) (x-4) <0
To solve the inequality \((x+1)(x-2)(x-4) < 0\), follow these steps: 1. **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 \(x = 4\). 2. **Determine the intervals:** The critical points divide the number line into the following intervals: - \((-\infty, -1)\) - \((-1, 2)\) - \((2, 4)\) - \((4, \infty)\) 3. **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.) 4. **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. 5. **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: \[ (-\infty, -1) \cup (2, 4) \]