Question

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

Answer 1

Ответ:

Объяснение:

Решение в приложении

Answer 2

To solve the inequality $(x-2)(x+3)(x-1)^2 \geq 0$, we can use the concept of intervals and test values. Here's how you can approach it:

1. Find the critical points by setting each factor equal to zero and solving for $x$: $(x-2) = 0$ gives $x = 2$ $(x+3) = 0$ gives $x = -3$ $(x-1) = 0$ gives $x = 1$

2. Plot these critical points on a number line: We have -3, 1, and 2.

3. Select test values from each interval: Choose a value less than -3, a value between -3 and 1, a value between 1 and 2, and a value greater than 2.

   Let's choose -4, 0, 1.5, and 3 as test values.

4. Evaluate the inequality for each test value: For $x = -4$: $(-4-2)(-4+3)(-4-1)^2 = (-6)(-1)(9) = 54$, which is positive. $\Rightarrow$ The inequality is true for $x = -4$.

   For $x = 0$: $(0-2)(0+3)(0-1)^2 = (-2)(3)(1) = -6$, which is negative. $\Rightarrow$ The inequality is false for $x = 0$.

   For $x = 1.5$: $(1.5-2)(1.5+3)(1.5-1)^2 = (-0.5)(4.5)(0.5)^2 = -0.5625$, which is negative. $\Rightarrow$ The inequality is false for $x = 1.5$.

   For $x = 3$: $(3-2)(3+3)(3-1)^2 = (1)(6)(4) = 24$, which is positive. $\Rightarrow$ The inequality is true for $x = 3$.

5. Analyze the results: From the test values, we see that the inequality is true for $x < -3$ and $1 < x \leq 2$. Therefore, the solution to the inequality $(x-2)(x+3)(x-1)^2 \geq 0$ is: $-\infty < x \leq -3$ or $1 < x \leq 2$