Question

(2x - 3)(x + 5)(3x - 1)^3 < 0;

Answer 1

Ответ:

$x$ и $\frac 1 3 \le x < \frac 3 2$

Объяснение:

$(2x-3)(x+5)(3x-1)^3=2(x-\frac{3}{2})(x+5)*27(x-\frac 1 3)$

Равносильно $(x-\frac{3}{2})(x+5)(x-\frac 1 3)$

Если $x$, то это подходит

Если $-5 \le x < \frac 1 3$, не подходит

Если $\frac 1 3 \le x < \frac 3 2$ подходит

Если $\frac 3 2 \le x$ не подходит

Answer 2

To solve the inequality (2x - 3)(x + 5)(3x - 1)^3 < 0, you can use the concept of interval notation and the properties of polynomial functions. Here are the steps to find the solution:

1. Identify the critical points by setting each factor equal to zero and solving for x: a) 2x - 3 = 0 b) x + 5 = 0 c) 3x - 1 = 0

Solve each of these equations: a) 2x - 3 = 0 2x = 3 x = 3/2

b) x + 5 = 0 x = -5

c) 3x - 1 = 0 3x = 1 x = 1/3

These values of x are the critical points.

2. Create intervals on the number line using these critical points: -∞ < -5 < 1/3 < 3/2 < ∞

3. Test each interval by picking a test point from each interval and evaluating the sign of the expression (2x - 3)(x + 5)(3x - 1)^3 in that interval. You can choose values such as x = -6, x = 0, x = 1, and x = 2 to test the intervals.

4. Analyze the signs of the expression in each interval: a) Interval (-∞, -5): Pick x = -6 (2(-6) - 3)(-6 + 5)(3(-6) - 1)^3 (-15)(-1)(-19)^3 = -15 * -1 * (-6859) = 102,885 The expression is positive in this interval.

   b) Interval (-5, 1/3): Pick x = 0 (2(0) - 3)(0 + 5)(3(0) - 1)^3 (-3)(5)(-1)^3 = -3 * 5 * (-1) = 15 The expression is positive in this interval.

   c) Interval (1/3, 3/2): Pick x = 1 (2(1) - 3)(1 + 5)(3(1) - 1)^3 (-1)(6)(2)^3 = -1 * 6 * 8 = -48 The expression is negative in this interval.

   d) Interval (3/2, ∞): Pick x = 2 (2(2) - 3)(2 + 5)(3(2) - 1)^3 (1)(7)(17)^3 = 1 * 7 * 4913 = 34,391 The expression is positive in this interval.

5. Based on the analysis of the signs in each interval, you can conclude the solution to the inequality: -∞ < x < 1/3 or 3/2 < x < ∞

So the solution to the inequality (2x - 3)(x + 5)(3x - 1)^3 < 0 is:

x in (-∞, 1/3) or (3/2, ∞)