Question

Решите уравнения f'(X)=0: 1) f(X)=3x^3-x 2)f(X)=9x^3+x

Answer 1

1. найдем производную. Она равна 9х²-1=0 отсюа х=±1/3

2. производная равна 27х²+1=0, Но сумма положительного и неотрицательного чисел не может быть равной нулю. Корней нет.

Answer 2

To find the critical points of the functions, we need to solve the equation f'(X) = 0 for each function.

1. f(X) = 3x^3 - x

Let's find the derivative of f(X) with respect to X:

f'(X) = d/dX (3x^3 - x) = 9x^2 - 1

Now, set f'(X) = 0 and solve for x:

9x^2 - 1 = 0

Adding 1 to both sides:

9x^2 = 1

Dividing both sides by 9:

x^2 = 1/9

Taking the square root of both sides:

x = ±√(1/9) = ±1/3

So, the critical points for the function f(X) = 3x^3 - x are x = 1/3 and x = -1/3.

2. f(X) = 9x^3 + x

Let's find the derivative of f(X) with respect to X:

f'(X) = d/dX (9x^3 + x) = 27x^2 + 1

Now, set f'(X) = 0 and solve for x:

27x^2 + 1 = 0

Subtracting 1 from both sides:

27x^2 = -1

Dividing both sides by 27:

x^2 = -1/27

This equation has no real solutions for x since the square of any real number cannot be negative. Therefore, the function f(X) = 9x^3 + x has no critical points.

To summarize:

1. For the function f(X) = 3x^3 - x, the critical points are x = 1/3 and x = -1/3.
2. The function f(X) = 9x^3 + x has no critical points.