Question

Y=6x^2-x^3 на [-1;6]

Answer 1

Решение задания смотри на фотографии

Answer 2

To find the extreme points of the function Y = 6x^2 - x^3 on the interval [-1, 6], we need to first find the critical points and then check the values of the function at these points and at the endpoints of the interval.

Step 1: Find the critical points Critical points occur when the derivative of the function is equal to zero or does not exist. Let's find the derivative of Y with respect to x:

Y = 6x^2 - x^3

dY/dx = d(6x^2)/dx - d(x^3)/dx dY/dx = 12x - 3x^2

Now, set the derivative equal to zero and solve for x:

12x - 3x^2 = 0

Factor out x:

x(12 - 3x) = 0

x = 0 or 12 - 3x = 0 x = 0 or 3x = 12 x = 0 or x = 4

Step 2: Check the values of Y at the critical points and the endpoints Now, we need to evaluate the function Y at the critical points and the endpoints of the interval [-1, 6]:

a) x = -1 Y(-1) = 6(-1)^2 - (-1)^3 Y(-1) = 6 - (-1) Y(-1) = 7

b) x = 0 (critical point) Y(0) = 6(0)^2 - (0)^3 Y(0) = 0

c) x = 4 (critical point) Y(4) = 6(4)^2 - (4)^3 Y(4) = 6(16) - 64 Y(4) = 96 - 64 Y(4) = 32

d) x = 6 Y(6) = 6(6)^2 - (6)^3 Y(6) = 6(36) - 216 Y(6) = 216 - 216 Y(6) = 0

Step 3: Determine the extreme points The extreme points (maximum or minimum) occur at critical points or endpoints where the function has the highest or lowest values.

In this case, the values of Y are as follows: Y(-1) = 7 Y(0) = 0 Y(4) = 32 Y(6) = 0

The maximum value of Y is 32, and it occurs at x = 4 (a critical point).

The minimum value of Y is 0, and it occurs at both x = 0 (a critical point) and x = 6 (an endpoint).

So, the extreme points of the function on the interval [-1, 6] are as follows: Maximum: (4, 32) Minimum: (0, 0) and (6, 0)