Question

F(x)=1/3x-x^3 нпйти макс и минимум​

Answer 1

f (x)=1/3x-x^2

0=1/3x-x^2

1/3x-x^2=0

x×(1/3-x)=0

x=0

1/3-x=0

x1=0

x2=1/3

Answer 2

To find the maximum and minimum of the function $F(x) = \frac{1}{3}x - x^3$, we need to find the critical points where the derivative of the function is equal to zero or undefined. We'll follow these steps:

1. Find the derivative of $F(x)$ with respect to $x$.
2. Set the derivative equal to zero and solve for $x$ to find the critical points.
3. Check the second derivative to determine if these critical points correspond to a maximum or minimum.

Step 1: Find the derivative of $F(x)$ with respect to $x$: $F'(x) = \frac{d}{dx}\left(\frac{1}{3}x - x^3\right)$ $F'(x) = \frac{1}{3} - 3x^2$

Step 2: Set the derivative equal to zero and solve for $x$ to find the critical points: $\frac{1}{3} - 3x^2 = 0$ $3x^2 = \frac{1}{3}$ $x^2 = \frac{1}{9}$ $x = \pm \frac{1}{3}$

Step 3: Check the second derivative to determine if these critical points correspond to a maximum or minimum: $F''(x) = \frac{d^2}{dx^2}\left(\frac{1}{3}x - x^3\right)$ $F''(x) = -6x$

Now, evaluate $F''(x)$ at the critical points: $F''\left(\frac{1}{3}\right) = -6 \times \frac{1}{3} = -2$ (negative, so it's a maximum) $F''\left(-\frac{1}{3}\right) = -6 \times \left(-\frac{1}{3}\right) = 2$ (positive, so it's a minimum)

Now we can conclude that:

• The maximum occurs at $x = \frac{1}{3}$ with a value of $F\left(\frac{1}{3}\right) = \frac{1}{27}$.
• The minimum occurs at $x = -\frac{1}{3}$ with a value of $F\left(-\frac{1}{3}\right) = -\frac{1}{27}$.