Let us analyze the mathematical statements from the image step by step.

First Statement

"На интервале $(\pi/2, 3\pi/2)$ найдется точка $c$, в которой производная функции $y = \cos x$ равна нулю."

• The derivative of $y = \cos x$ is $y' = -\sin x$.
• We are asked to check if $-\sin x = 0$ within the interval $(\pi/2, 3\pi/2)$.

Within this interval:

• The sine function, $\sin x$, equals zero at $x = \pi$, which lies in $(\pi/2, 3\pi/2)$.

Conclusion: This statement is true.

Second Statement

"Точка $x_0 = 0$ — точка строгого локального минимума функции $y = |x|$."

• The function $y = |x|$ is defined as: x, & x \geq 0 \\ -x, & x < 0 \end{cases}$$
• At $x_0 = 0$, the value of $y$ is $0$. The value increases as $x$ moves away from $0$ in either direction ($y > 0$ for $x \neq 0$).

Conclusion: $x_0 = 0$ is indeed a strict local minimum point for $y = |x|$. This statement is true.

Third Statement

"Поскольку функция $y = |x|$ в точке $x_0 = 0$ принимает экстремальное значение и имеет конечную производную $y'(x_0)$, то $y'(x_0) = 0$."

• For $y = |x|$, the derivative is: 1, & x > 0 \\ -1, & x < 0 \end{cases}$$
• At $x = 0$, the derivative does not exist because the left-hand derivative ($-1$) does not equal the right-hand derivative ($1$).

Conclusion: This statement is false, as the derivative at $x_0 = 0$ is not defined.

Fourth Statement

"На интервале $(0, 2)$ найдется точка $c$, в которой производная функции $y = x^3$ равна $y'(c) = -4$."

• The derivative of $y = x^3$ is $y' = 3x^2$.
• We are asked to solve $3x^2 = -4$ for some $x \in (0, 2)$.
• Since $3x^2 \geq 0$ for all $x$, it is impossible for $3x^2$ to equal $-4$.

Conclusion: This statement is false.

Fifth Statement

"На интервале $(0, 2)$ найдется точка $c$, в которой для производных функций $f(x) = x^4$ и $g(x) = x$ выполняется соотношение $f'(c) = 8g'(c)$."

• The derivatives are: $f'(x) = 4x^3, \quad g'(x) = 1.$
• Substituting into the equation $f'(c) = 8g'(c)$, we get: $4c^3 = 8 \implies c^3 = 2 \implies c = \sqrt[3]{2}.$
• Since $c = \sqrt[3]{2} \approx 1.26$, which lies in the interval $(0, 2)$, such a point $c$ exists.

Conclusion: This statement is true.

Final Answer

The correct statements are:

1. The first statement.
2. The second statement.
3. The fifth statement.

Would you like a detailed explanation of any specific part? Let me know!

5 Related Questions

1. What is the process for finding critical points of a function in a given interval?
2. How do we determine if a point is a local minimum or maximum using derivatives?
3. Why does the derivative of $y = |x|$ not exist at $x = 0$?
4. How do we solve equations involving cubic roots, such as $c^3 = 2$?
5. What are the conditions for applying the mean value theorem for derivatives?

Tip

When analyzing derivatives, always check for both existence and continuity to avoid incorrect conclusions!