Question

Y= 2 +15x +2х*корень из х Найти точку max

Answer 1

Ответ:

Нет точки максимума

Объяснение:

Рассмотрим функцию

$\displaystyle y=2+15*x+2*x*\sqrt{x} =2+15*x+2*x^{\frac{3}{2} }$

Так как в составе функции участвует квадратный корень, то область определений функции: x≥0, то есть D(y)=[0; +∞).

Чтобы найти экстремумы (локальные минимумы и максимумы) будем исследовать функцию с помощью производной функции. Вычислим производную функции:

$\displaystyle y'=(2+15*x+2*x^{\frac{3}{2}})'=(2)'+15*(x)'+2*(x^{\frac{3}{2}})'=\\\\=0+15*1+2*\frac{3}{2}}*x^{\frac{3}{2}-1}=15+3*\sqrt{x}$

Так как $\displaystyle \sqrt{x}\geq 0$ , то

$\displaystyle y'=15+3*\sqrt{x}\geq 15>0$

для любого x∈D(y). Это означает, что данная функция монотонно возрастает в D(y). Отсюда следует, что у функции нет точки максимума.

Так как функция монотонно возрастает в D(y), то минимальное значение в D(y)=[0; +∞) принимает при x=0: y(0)=2.

Answer 2

To find the maximum point of the function Y = 2 + 15x + 2x * √x, you need to find the critical points and then determine which one corresponds to a maximum.

Step 1: Find the derivative of Y with respect to x:

Y = 2 + 15x + 2x * √x

To find the derivative, we'll use the product rule for the term "2x * √x":

d/dx [2x * √x] = 2 * √x + 2x * (1/2) * x^(-1/2) = 2 * √x + x^(3/2)

So, the derivative of Y with respect to x is:

dY/dx = 15 + 2 * √x + x^(3/2)

Step 2: Find the critical points by setting the derivative equal to zero and solving for x:

15 + 2 * √x + x^(3/2) = 0

This equation may not have a simple algebraic solution, but we can use numerical methods to find approximate values for x. One common numerical method is the Newton-Raphson method. For this purpose, let's start with an initial guess, say x0 = 1.

Step 3: Use the Newton-Raphson method to find the critical points:

x1 = x0 - [f(x0) / f'(x0)] x1 = 1 - [(15 + 2 * √1 + 1^(3/2)) / (2 * √1 + 1^(3/2))] x1 ≈ 0.757

Now we have an approximate critical point x1 ≈ 0.757.

Step 4: Confirm whether this point is a maximum or minimum by analyzing the second derivative.

To do that, find the second derivative of Y with respect to x:

d^2Y/dx^2 = d/dx [15 + 2 * √x + x^(3/2)] d^2Y/dx^2 = 2 * (1/2) * x^(-1/2) + (3/2) * x^(1/2) d^2Y/dx^2 = x^(-1/2) + (3/2) * x^(1/2)

Now, substitute the critical point x1 ≈ 0.757 into the second derivative:

d^2Y/dx^2 ≈ 0.757^(-1/2) + (3/2) * 0.757^(1/2) ≈ 2.82

Since the second derivative is positive (approximately 2.82), the point x1 ≈ 0.757 corresponds to a minimum, not a maximum.

Step 5: To find the maximum point, we need to consider the behavior of the function as x approaches infinity and negative infinity. The term with the highest degree in x is 15x, which dominates as x goes to positive or negative infinity.

As x goes to positive infinity, the function Y = 2 + 15x + 2x * √x will also go to positive infinity. So, there is no maximum point.

As x goes to negative infinity, the function Y will go to negative infinity. Again, there is no maximum point in this case either.

Therefore, the function has no maximum point. It only has a minimum point at x ≈ 0.757.