Question

У=2(х+1) аргумент - 3<х<3

Answer 1

Вот ответ
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Answer 2

To find the range of the function $У=2(х+1)$ when the argument $х$ is within the interval $-3 < х < 3$, we need to evaluate the function at the boundary points of the interval and determine the minimum and maximum values.

Step 1: Find $У$ at the boundary points of the interval $-3 < х < 3$:

• At $х = -3$: $У = 2(-3 + 1) = 2(-2) = -4$
• At $х = 3$: $У = 2(3 + 1) = 2(4) = 8$

Step 2: Determine the minimum and maximum values of $У$ within the interval: Since the coefficient of $х$ in the function is positive (2), the function will increase as $х$ increases. Therefore, the minimum value of $У$ occurs at the lower boundary, $х = -3$, and the maximum value of $У$ occurs at the upper boundary, $х = 3$.

So, the range of the function $У=2(х+1)$ when the argument $х$ is within the interval $-3 < х < 3$ is $-4 < У < 8$.