Question

Дам 30 баллов! Определите точку максимума функции y=(x-4)^2 *(x-1) Пожалуйста с подробным решением!!!

Answer 1

Решение:
Раскрываем скобки и приводим подобные слагаемые: 
у=x^3-9x^2+24x-16 
Производная у = 3х^2-18x+24=0 
x^2-6x+8=0
х=4
х=2 - точка максимум
В этой точке производная функции меняет знак с "+" на "-".
у (2) = (-2)^2*1=4
Ответ: (2;4)

Answer 2

Finding the Maximum Point of the Function

To find the maximum point of the function y = (x-4)^2 * (x-1), we can use calculus. The maximum point occurs when the derivative of the function is equal to zero.

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

y = (x-4)^2 * (x-1)

To simplify the calculation, we can expand the function:

y = (x^2 - 5x + 4) * (x-1)

Expanding further:

y = x^3 - 6x^2 + 9x - 4

Now, let's find the derivative of the function:

dy/dx = 3x^2 - 12x + 9

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

3x^2 - 12x + 9 = 0

We can factor the equation:

3(x^2 - 4x + 3) = 0

Factoring further:

3(x-1)(x-3) = 0

Setting each factor equal to zero:

x-1 = 0 or x-3 = 0

Solving for x:

x = 1 or x = 3

Now, we have two critical points: x = 1 and x = 3. To determine whether these points correspond to a maximum or minimum, we can use the second derivative test.

Taking the second derivative of the function:

d^2y/dx^2 = 6x - 12

Substituting x = 1 and x = 3 into the second derivative:

d^2y/dx^2 (x=1) = 6(1) - 12 = -6

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

Since the second derivative is negative at x = 1 and positive at x = 3, we can conclude that x = 1 corresponds to a maximum point and x = 3 corresponds to a minimum point.

To find the corresponding y-values, we substitute x = 1 and x = 3 back into the original function:

y(x=1) = (1-4)^2 * (1-1) = 9 * 0 = 0

y(x=3) = (3-4)^2 * (3-1) = 1 * 2 = 2

Therefore, the maximum point of the function y = (x-4)^2 * (x-1) is at (1, 0).

Conclusion

The maximum point of the function y = (x-4)^2 * (x-1) is at (1, 0).