Question

Помогите решить :Найдите наибольшее и наименьшее значение функции y = -x^3 +3x^2 +4 на отрезке (-3:3)

Answer 1

y'=-3x^2+6x

y'=0

-3x^2+6x=0

3x(2-x)=0

x=0

x=2

попадают 

Теперь на концах вычеслим значения  

f(-3)=-3^3+3*9+4<4

f(3)=27+3*9+4<58

f(0)=4

f(2)=8+12+4=24

 

 то есть наибольшее меньше 58, наименьшее меньше 4

Answer 2

Finding the Maximum and Minimum Values of a Function

To find the maximum and minimum values of the function y = -x^3 + 3x^2 + 4 on the interval (-3, 3), we can use the first and second derivative tests to locate the critical points and determine whether they correspond to maximum or minimum values.

First, let's find the critical points by finding the derivative of the function and setting it equal to zero to solve for the critical x-values.

The derivative of the function is: y' = -3x^2 + 6x

Setting y' equal to zero and solving for x: -3x^2 + 6x = 0 x(-3x + 6) = 0 x = 0 or x = 2

Now, we need to determine whether these critical points correspond to maximum or minimum values by using the second derivative test.

The second derivative of the function is: y'' = -6x + 6

Substituting the critical points x = 0 and x = 2 into y'': y''(0) = 6 y''(2) = -6

Since y''(0) > 0, the function has a local minimum at x = 0. And since y''(2) < 0, the function has a local maximum at x = 2.

Now that we have identified the critical points and their corresponding nature, we can evaluate the function at these points to find the maximum and minimum values.

Substituting x = 0 and x = 2 into the original function: y(0) = 4 y(2) = 4

Therefore, the maximum and minimum values of the function y = -x^3 + 3x^2 + 4 on the interval (-3, 3) are: Maximum value: 4 Minimum value: 4