Question

Y=x³-6x²+9 найти экстремумы функции

Answer 1

Ответ:

Пошаговое объяснение:

находим критические точки

находим производную

3x²-12x

приравняли к 0

x1=0 x2=4

производная при переходе через точку x=0 меняет знак с + на -

значит в этой точке имеем максимум.

в точке x=4 производная меняет с -на+ и имеется минимум

y(0)=9

y(4)=64+9-96=-23

Answer 2

Finding the Extrema of the Function Y = x³ - 6x² + 9

To find the extrema of the function Y = x³ - 6x² + 9, we need to determine the critical points where the derivative of the function is equal to zero or does not exist. The critical points correspond to the potential extrema of the function.

Let's find the derivative of the function Y = x³ - 6x² + 9:

Y' = 3x² - 12x

To find the critical points, we set the derivative equal to zero and solve for x:

3x² - 12x = 0

Factoring out x, we get:

x(3x - 12) = 0

Setting each factor equal to zero, we find two critical points:

x = 0 and x = 4

Now, we need to determine whether these critical points correspond to a maximum or minimum. We can do this by analyzing the second derivative of the function.

Let's find the second derivative of the function Y = x³ - 6x² + 9:

Y'' = 6x - 12

Substituting the critical points into the second derivative, we get:

Y''(0) = 6(0) - 12 = -12 Y''(4) = 6(4) - 12 = 12

Since the second derivative is negative at x = 0 and positive at x = 4, we can conclude that x = 0 corresponds to a maximum and x = 4 corresponds to a minimum.

Therefore, the extrema of the function Y = x³ - 6x² + 9 are as follows:

- Maximum at x = 0 - Minimum at x = 4

Please note that the above information is based on the provided function and its derivatives.