Question

Найти максимум и минимум функции 2+sinx

Answer 1

Ответ на фото ........

Answer 2

To find the maximum and minimum of the function $f(x) = 2 + \sin(x)$, we need to analyze its behavior and find critical points.

Step 1: Find the derivative of $f(x)$: $f'(x) = \frac{d}{dx} (2 + \sin(x)) = \cos(x)$

Step 2: Find critical points by setting the derivative to zero and solving for $x$: $\cos(x) = 0$

The critical points occur when $x$ is an odd multiple of $\pi/2$: $x = \frac{\pi}{2} + k\pi, \text{ where } k \in \mathbb{Z}$

Step 3: Determine the nature of the critical points by looking at the second derivative: $f''(x) = \frac{d}{dx} (\cos(x)) = -\sin(x)$

Step 4: Evaluate the second derivative at the critical points: $f''\left(\frac{\pi}{2} + k\pi\right) = -\sin\left(\frac{\pi}{2} + k\pi\right)$

Now, we know that $\sin\left(\frac{\pi}{2}\right) = 1$ and $\sin\left(\frac{3\pi}{2}\right) = -1$. Since $\sin$ is periodic with a period of $2\pi$, it alternates between 1 and -1.

Thus, when $k$ is even, $f''\left(\frac{\pi}{2} + k\pi\right) = -1$, and when $k$ is odd, $f''\left(\frac{\pi}{2} + k\pi\right) = 1$.

Step 5: Identify maximum and minimum points:

• Maximum points occur where the second derivative changes from negative to positive, i.e., when $k$ changes from even to odd.
• Minimum points occur where the second derivative changes from positive to negative, i.e., when $k$ changes from odd to even.

Step 6: Evaluate $f(x)$ at the critical points to find the maximum and minimum values.

Let's find the maximum and minimum values:

1. Maximum points: When $k$ is even, i.e., $k = 0, 2, -2, \ldots$

Maximum value: $f\left(\frac{\pi}{2} + k\pi\right) = 2 + \sin\left(\frac{\pi}{2} + k\pi\right) = 2 + 1 = 3$

2. Minimum points: When $k$ is odd, i.e., $k = 1, -1, 3, \ldots$

Minimum value: $f\left(\frac{\pi}{2} + k\pi\right) = 2 + \sin\left(\frac{\pi}{2} + k\pi\right) = 2 + (-1) = 1$

Therefore, the maximum value of the function $2 + \sin(x)$ is 3, and the minimum value is 1.