Question

Неравенство 2^(9x-x^3)<1

Answer 1

Ответ:

Пошаговое объяснение:

2^(9x-x^3)<2^0,   9x-x^3<0,   x(9-x^2)<0,   x(3-x)(3+x)<0,  решаем методом интервалов,   ______(-3)_______(0)_______(3)______

                             +                     -                   +               -                            

нам нужно  < 0,   ответ:  (-3;0)  и  ((3;+~)

Answer 2

To solve the inequality 2^(9x - x^3) < 1, we can take the logarithm of both sides. However, since the base of the exponent is 2, we should take the logarithm base 2 of both sides. The inequality becomes:

log2(2^(9x - x^3)) < log2(1)

Using the logarithmic property log_b(a^c) = c * log_b(a), we simplify the inequality:

(9x - x^3) * log2(2) < 0

Simplifying further, we have:

9x - x^3 < 0

Now, let's solve this inequality. To do that, we can factor it as follows:

x(x^2 - 9) < 0

The roots of the equation x^2 - 9 = 0 are x = -3 and x = 3. These points divide the number line into three intervals: (-∞, -3), (-3, 3), and (3, ∞).

Now, we can test each interval by choosing a test point within it and checking if it satisfies the inequality.

For the interval (-∞, -3), we can choose x = -4:

(-4)(-4^2 - 9) < 0 (-4)(16 - 9) < 0 (-4)(7) < 0 -28 < 0 (True)

For the interval (-3, 3), we can choose x = 0:

(0)(0^2 - 9) < 0 0 < 0 (False)

For the interval (3, ∞), we can choose x = 4:

(4)(4^2 - 9) < 0 (4)(16 - 9) < 0 (4)(7) < 0 28 < 0 (False)

Therefore, the solution to the inequality 2^(9x - x^3) < 1 is x ∈ (-∞, -3) ∪ (3, ∞).