Question

Не как не могу сделать задание по алгебре.. Найдите точки экстремума функции и определите их характер: y= - x3/3 - 2x2 + 3

Answer 1

f'(x)=-x^2-4x=0

x1=-4

x2=0

от -∞ до -4 f'(x)<0 функция убывает

от -4 до 0 f'(x)>0 функция возрастает

от 0 до +∞f'(x)<0 функция убывает

Значит в значении x=-4 минимум функции

в  x2=0 максимум

 

 

Answer 2

Finding the Extrema of the Function

To find the extrema of the function y = -x^3/3 - 2x^2 + 3, we need to determine the critical points where the derivative of the function is equal to zero or does not exist. The critical points will help us identify the extrema and their nature.

Let's start by finding the derivative of the function:

y' = -x^2 - 4x

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

-x^2 - 4x = 0

Factoring out x, we get:

x(-x - 4) = 0

This equation is satisfied when either x = 0 or -x - 4 = 0. Solving for x in the second equation, we find:

-x - 4 = 0 -x = 4 x = -4

So, the critical points are x = 0 and x = -4.

Determining the Nature of the Extrema

To determine the nature of the extrema at these critical points, we can use the second derivative test. The second derivative of the function is:

y'' = -2x - 4

Let's evaluate the second derivative at the critical points:

For x = 0: y''(0) = -2(0) - 4 = -4

For x = -4: y''(-4) = -2(-4) - 4 = 8 - 4 = 4

According to the second derivative test: - If the second derivative is positive at a critical point, the function has a local minimum at that point. - If the second derivative is negative at a critical point, the function has a local maximum at that point. - If the second derivative is zero or does not exist at a critical point, the test is inconclusive.

Based on the second derivative test, we can conclude the following:

- At x = 0, the second derivative is negative (-4), indicating a local maximum at that point. - At x = -4, the second derivative is positive (4), indicating a local minimum at that point.

Therefore, the function y = -x^3/3 - 2x^2 + 3 has a local maximum at x = 0 and a local minimum at x = -4.

I hope this helps! Let me know if you have any further questions.