Solution to problem 15.1.9 from the collection of Kepe O.E.

Problem 15.1.9 from the collection of Kepe O.?. refers to the section of mathematical analysis and has the following condition:

"Prove that the function $f(x) = x^3 - 3x^2 + 2$ is monotonically increasing on the interval $[-1,2]$."

To solve this problem, it is necessary to analyze the derivative of the function $f(x)$ on the specified interval. If the derivative of the function $f'(x)$ is positive on the entire interval $[-1,2]$, then this will mean that the function $f(x)$ increases monotonically on this interval.

Let's calculate the derivative of the function $f(x)$:

$f'(x) = 3x^2 - 6x$

Now let’s find the roots of this derivative, equating it to zero:

$f'(x) = 0$

$3x^2 - 6x = 0$

$x(3x - 6) = 0$

$x_1 = 0, x_2 = 2$

We get two points where the derivative of the function is zero. There are three intervals left on the segment $[-1,2]$:

1. $[-1,0)$
2. $(0,2)$
3. $[2,+\infty)$

From the table of signs of the derivative of a function one can see that on the interval $[-1,0)$ the derivative is negative, on the interval $(0,2)$ the derivative is positive, and on the interval $[2,+\infty)$ the derivative is also positive.

Thus, on the entire interval $[-1,2]$ the derivative of the function $f(x)$ is positive, which means that the function $f(x)$ increases monotonically on this interval. Problem 15.1.9 from the collection of Kepe O.?. solved.

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Problem 15.1.9 from the collection of Kepe O.?. is that you need to find the value of the integral of the product of two functions f(x) and g(x) on a given interval [a, b]. To solve this problem, it is necessary to use methods of mathematical analysis, such as the method of rectangles, the method of trapezoids or the Simpson method.

First, you need to analyze the functions f(x) and g(x) specified in the problem statement and determine which integration method is best to use in this case. Then you need to apply the selected method to calculate the value of the integral.

Solving Problem 15.1.9 can be quite complex, so it is important to have a good knowledge of mathematical analysis and the ability to apply integration methods in various situations.

Solution to problem 15.1.9 from the collection of Kepe O.?. requires determining the work of gravity during the first half of the oscillation period of a body weighing 0.1 kg suspended at the end of an unstretched spring with a stiffness coefficient c = 50 N/m.

It is known that when a body is released without an initial speed, the spring begins to oscillate around its equilibrium position. The oscillation period of the spring can be calculated using the formula T = 2π√(m/c), where m is the body mass, c is the spring stiffness coefficient.

The first half of the oscillation period corresponds to the moment when the body passes the equilibrium position and begins to move in the opposite direction. In this section of the trajectory of motion, the body slows down under the action of gravity directed downwards, and the work of this force is calculated by the formula A = mgh, where g is the acceleration of free fall, h is the height to which the body has risen relative to the equilibrium position of the spring.

The lifting height of a body can be found from the law of conservation of energy of a mechanical system, including a body and a spring. The initial potential energy of the system is 0, since the body was released without an initial velocity. This means that the total mechanical energy of the system at the initial moment of time is equal to the kinetic energy of the body, which is also equal to 0. At the moment the body passes through the equilibrium position, the kinetic energy of the body is also equal to 0, and the potential energy of the system is maximum and equal to 0.5kh^2, where k is spring stiffness coefficient, h - maximum displacement of the spring relative to the equilibrium position.

According to the law of conservation of energy of a mechanical system, the maximum displacement of the spring can be found from the equation 0.5kh^2 = mgh, from where h = √(2mg/k).

Thus, the work done by gravity during the first half of the oscillation period is equal to A = mgh = mg√(2mg/k). Substituting the known values, we obtain A = 9.62 • 10^-3 J.

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