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Задача Задача №2626209
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Аноним
13 января 2025 22:19 · публичный вопрос
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Аноним

(x ^ 2 - 1) * (x + 3) ^ 2 >= 0

Ответ нейросети
13 января 2025 22:19

To solve the inequality \((x^2 - 1)(x + 3)^2 \geq 0\), we will first find the critical points by identifying the values of \(x\) that make the expression equal to zero or undefined.

Step 1: Find the critical points

  1. Set each factor in the expression to zero:
    • For \(x^2 - 1 = 0\): \[ x^2 = 1 \implies x = 1 \text{ or } x = -1 \]

    • For \((x + 3)^2 = 0\): \[ x + 3 = 0 \implies x = -3 \]

So, the critical points are \(x = -3\), \(x = -1\), and \(x = 1\).

Step 2: Determine the sign of the expression in the intervals

The critical points divide the number line into four intervals:

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

We will test a point in each interval to determine the sign of the expression.

  • Interval \( (-\infty, -3) \): Let's pick \(x = -4\) \[ (-4)^2 - 1 = 16 - 1 = 15 > 0 \] \[ (-4 + 3)^2 = (-1)^2 = 1 > 0 \quad \Rightarrow \quad (15)(1) > 0 \] Thus, the expression is positive in this interval.

  • Interval \( (-3, -1) \): Let's pick \(x = -2\) \[ (-2)^2 - 1 = 4 - 1 = 3 > 0 \] \[ (-2 + 3)^2 = (1)^2 = 1 > 0 \quad \Rightarrow \quad (3)(1) > 0 \] Thus, the expression is positive in this interval.

  • Interval \( (-1, 1) \): Let's pick \(x = 0\) \[ (0)^2 - 1 = 0 - 1 = -1 < 0 \] \[ (0 + 3)^2 = (3)^2 = 9 > 0 \quad \Rightarrow \quad (-1)(9) < 0 \] Thus, the expression is negative in this interval.

  • Interval \( (1, \infty) \): Let's pick \(x = 2\) \[ (2)^2 - 1 = 4 - 1 = 3 > 0 \] \[ (2 + 3)^2 = (5)^2 = 25 > 0 \quad \Rightarrow \quad (3)(25) > 0 \] Thus, the expression is positive in this interval.

Step 3: Include the critical points

Next, we check the sign at the critical points:

  • At \(x = -3\): \[ (x^2 - 1)(x + 3)^2 = (9 - 1)(0) = 0 \]

  • At \(x = -1\): \[ (x^2 - 1)(x + 3)^2 = (1 - 1)(4) = 0 \]

  • At \(x = 1\): \[ (x^2 - 1)(x + 3)^2 = (1 - 1)(16) = 0 \]

Step 4: Combine the results

The expression \((x^2 - 1)(x + 3)^2 \geq 0\) is satisfied in:

  • The intervals \( (-\infty, -3) \), \( (-3, -1) \), and \( (1, \infty) \)
  • At the points \(x = -3\), \(x = -1\), and \(x = 1\)

Final Solution

Combining all the information, the solution to the inequality \((x^2 - 1)(x + 3)^2 \geq 0\) is: \[ (-\infty, -3] \cup [-1, 1] \cup [1, \infty) \]