Question

(4-х)(3х-1)(х+8) ≤ 0​

Answer 1

Ответ:

[-8;$\frac{1}{3}$]U[4;+бесконечность)

Пошаговое объяснение:

Answer 2

To solve the inequality $(4-х)(3х-1)(х+8) ≤ 0$, we can follow these steps:

1. Find the critical points by setting each factor equal to zero and solving for $x$:

   • $4 - x = 0$ ⇒ $x = 4$
   • $3x - 1 = 0$ ⇒ $x = \frac{1}{3}$
   • $x + 8 = 0$ ⇒ $x = -8$

2. Create intervals based on these critical points: $-\infty < -8 < \frac{1}{3} < 4 < \infty$.

3. Test a value from each interval in the original inequality to determine the sign of the expression:

   • For $x = -9$: $(4 - (-9))(3(-9) - 1)(-9 + 8) > 0$ (positive)
   • For $x = 0$: $(4 - 0)(3(0) - 1)(0 + 8) < 0$ (negative)
   • For $x = \frac{1}{2}$: $(4 - \frac{1}{2})(3(\frac{1}{2}) - 1)(\frac{1}{2} + 8) < 0$ (negative)
   • For $x = 5$: $(4 - 5)(3(5) - 1)(5 + 8) > 0$ (positive)

Based on these test values, we can see that the expression is negative in the interval $\frac{1}{3} < x < 4$ and positive outside of this interval.

So, the solution to the inequality $(4-х)(3х-1)(х+8) ≤ 0$ is: $\frac{1}{3} \leq x \leq 4$