Question

Y=√(−79−18x−x2)найти точку максимума функцииy​

Answer 1

$y = \sqrt{ - 79 - 18x - {x}^{2} }$

у' =

$- \frac{1}{2 \sqrt{ - {x}^{2} - 18x - 79} } \times ( - 2x - 18) = - \frac{ - 2(x + 9)}{2 \sqrt{ - {x}^{2} - 18x - 79} } = \frac{x + 9}{\sqrt{ - {x}^{2} - 18x - 79} }$

Приравняем производную к нулю :

$\frac{x + 9}{\sqrt{ - {x}^{2} - 18x - 79} } = 0$

ОДЗ :

-х²-18х-79 > 0

х²+18х+79 < 0

(х+9+√2)(х+9-√2) < 0

х € (-9-√2 ; -9+√2)≈ (-10.4 ; -7.6)

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х+9 = 0

х = -9 - Подходит под ОДЗ :

Подставим в функцию :

$y = \sqrt{ - 79 - 18 \times ( - 9) - { (- 9)}^{2} } = \sqrt{2}$

Ответ : √2

Answer 2

Ответ:

$(-9;\sqrt{2})$

Пошаговое объяснение:

Для того чтобы найти экстремумы, нужно решить уравнение:

$\frac{d}{dx} f(x)=0$ - Производная равна 0, а корни этого уравнения будут экстремумами данной функции:

$\frac{d}{dx} f(x)=\frac{-x-9}{\sqrt{-x^{2}+(-18x-79) }} =0$

Корни этого уравнения:

$x_{1} =-9$

Answer 3

To find the maximum point of the function $y = \sqrt{-79 - 18x - x^2}$, we need to determine the critical points and analyze the behavior of the function around those points.

First, let's differentiate the function with respect to $x$ to find the critical points:

$y = \sqrt{-79 - 18x - x^2}$ $y^2 = -79 - 18x - x^2$ $y^2 = -(x^2 + 18x + 79)$ $\frac{dy^2}{dx} = -(2x + 18)$

Setting $\frac{dy^2}{dx} = 0$, we can find the critical point:

$-(2x + 18) = 0$ $2x + 18 = 0$ $2x = -18$ $x = -9$

Now we need to determine the nature of the critical point. To do this, we can use the second derivative test. We differentiate $\frac{dy^2}{dx}$ again:

$\frac{d^2y^2}{dx^2} = -2$

Since the second derivative is negative ($-2$), the critical point at $x = -9$ corresponds to a maximum.

To find the maximum point, substitute the critical point into the original function:

$y = \sqrt{-79 - 18(-9) - (-9)^2}$ $y = \sqrt{-79 + 162 - 81}$ $y = \sqrt{2}$

Therefore, the maximum point of the function $y = \sqrt{-79 - 18x - x^2}$ is $(-9, \sqrt{2})$.