Question

Решить уравнение f(x)=12x-x^3 f'(x)=0

Answer 1

f’(x)=12-3x^2=0

12=3x^2

x^2=12/3

x^2=4

x=±2

Answer 2

To solve the equation f'(x) = 0 for the function f(x) = 12x - x^3, we need to find the critical points of the function. Critical points occur where the derivative is equal to zero or is undefined.

Let's start by finding the derivative of f(x) with respect to x:

f(x) = 12x - x^3

To find f'(x), we'll use the power rule for derivatives:

f'(x) = d/dx (12x - x^3) f'(x) = 12 - 3x^2

Now, set f'(x) equal to zero and solve for x:

12 - 3x^2 = 0

To solve for x, move the constant term to the other side:

3x^2 = 12

Now, divide both sides by 3:

x^2 = 4

To solve for x, take the square root of both sides:

x = ±√4

x = ±2

So, we have two critical points: x = 2 and x = -2.

To verify that these are indeed critical points, we can use the second derivative test. The second derivative of f(x) is:

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

Now, evaluate f''(x) at x = 2 and x = -2:

f''(2) = -6(2) = -12 f''(-2) = -6(-2) = 12

Since f''(2) < 0 and f''(-2) > 0, we can conclude that x = 2 is a local maximum and x = -2 is a local minimum.

Thus, the critical points of the function f(x) = 12x - x^3 occur at x = 2 (local maximum) and x = -2 (local minimum).