Question

Знайдіть екстремуми функції у = 2х^3- 3х^2;

Answer 1

Ответ:

Объяснение:

y=2х^3-3x^2

1)находим производную функции

y'= 2*3x^2 - 3*2x^1 = 6x^2-6x

2) приравниваем производную к нулю

y'=0

6x^2-6x = 0

6x(x-1) = 0

6x = 0    или  х-1=0

х=0                х=1

3)находим значение функции в корнях уравнения(найденных значениях х)

y(0) = 2*0^3-3*0^2=0-0=0                  y(0) больше, чем y(1)

y(1) = 2*1^3-3*1^2=2-3=-1

y max=y(0)=0

Answer 2

Finding Extremes of the Function ????(????) = 2????^3 - 3????^2

To find the extremes of the function ????(????) = 2????^3 - 3????^2, we can use calculus to find the critical points and then determine whether they correspond to a maximum or minimum.

The first step is to find the derivative of the function ????(????) with respect to ????. The critical points occur where the derivative is equal to zero or is undefined.

The derivative of ????(????) = 2????^3 - 3????^2 is: ????'(????) = 6????^2 - 6????

To find the critical points, we set the derivative equal to zero and solve for ????: 6????^2 - 6???? = 0 6????(???? - 1) = 0 ???? = 0, ???? = 1

Now, we need to determine whether these critical points correspond to a maximum or minimum. We can use the second derivative test or evaluate the function at these points to find the nature of the extremes.

Let's evaluate the function ????(????) at the critical points ???? = 0 and ???? = 1:

For ???? = 0: ????(0) = 2(0)^3 - 3(0)^2 = 0 For ???? = 1: ????(1) = 2(1)^3 - 3(1)^2 = -1

From the evaluations, we can see that ????(0) = 0 and ????(1) = -1.

Therefore, the minimum value of the function ????(????) = 2????^3 - 3????^2 is 0, and it occurs at ???? = 0.

Minimum Value: ????(0) = 0 at ???? = 0

This concludes the analysis of the extremes of the function ????(????) = 2????^3 - 3????^2. If you have any further questions or need additional assistance, feel free to ask!