Question

H(x) =3x-x^2-2 Промежутки знакопостоянства​

Answer 1

Ответ:

Промежутки знакопостоянства.

Answer 2

To determine the intervals of sign constancy for the function H(x) = 3x - x^2 - 2, we need to find the values of x where the function changes its sign.

First, let's find the critical points of the function by setting H(x) = 0 and solving for x:

3x - x^2 - 2 = 0

Rearranging the equation:

x^2 - 3x + 2 = 0

Factoring the quadratic equation:

(x - 1)(x - 2) = 0

So the critical points are x = 1 and x = 2.

Now, we'll check the sign of the function in the intervals between and outside of these critical points.

Interval 1: (-∞, 1) We can choose any value in this interval, such as x = 0, and substitute it into the function: H(0) = 3(0) - (0)^2 - 2 = -2

Interval 2: (1, 2) Again, we can choose a value within this interval, like x = 1.5: H(1.5) = 3(1.5) - (1.5)^2 - 2 = -1.75

Interval 3: (2, ∞) Using x = 3 as an example: H(3) = 3(3) - (3)^2 - 2 = -8

Summarizing the sign of H(x) in each interval:

Interval 1: (-∞, 1) -> H(x) is negative (-) Interval 2: (1, 2) -> H(x) is negative (-) Interval 3: (2, ∞) -> H(x) is negative (-)

Therefore, the function H(x) = 3x - x^2 - 2 is negative (-) for all x values in the intervals (-∞, 1) and (1, 2), as well as for x values in the interval (2, ∞).