Question

Найдите точку минимума функции . пример?!)

Answer 1

Первая производная в критических точках функции = 0
напр.
у = х² - 3х + 6
у' = 2x - 3 = 0
x = 3/2
y' < 0,  x ∈ (-∞;  3/2)   ⇒ функция убывает
y' > 0,  x ∈ (3/2; +∞)   ⇒ функция возрастает
Вершина параболы в точке с абсциссой  x = 3/2 -минимальная

Действительно,  у = х² - 3х + 6 - парабола, оси направлены вверх и вершина параболы в точке с абсциссой x = 3/2

Answer 2

Finding the Minimum Point of a Function

To find the minimum point of a function, we need to identify the x-value(s) where the function reaches its lowest point. Let's consider an example to illustrate this.

Suppose we have the function f(x) = x^2 + 8x - 7. To find the minimum point of this function, we can use calculus by taking the derivative of the function and setting it equal to zero. Solving for x will give us the x-coordinate of the minimum point.

Taking the derivative of f(x) with respect to x, we get:

f'(x) = 2x + 8

Setting f'(x) equal to zero and solving for x:

2x + 8 = 0 2x = -8 x = -4

So, the x-coordinate of the minimum point is x = -4. To find the corresponding y-coordinate, we substitute this value back into the original function:

f(-4) = (-4)^2 + 8(-4) - 7 f(-4) = 16 - 32 - 7 f(-4) = -23

Therefore, the minimum point of the function f(x) = x^2 + 8x - 7 is (-4, -23).

Additional Information

It's worth noting that there are other methods to find the minimum point of a function, such as using optimization algorithms like gradient descent or Newton's method. However, in this example, we used calculus to find the minimum point.

If you have a different function in mind or need further assistance, please let me know!