Question

F(x)=sqrt x^3-3x+11. [-2;13]

Answer 1

F(x)=√(x^3-3x+11).              [-2;13]
f'(x) = 1/ (2√(x³-3x+11)) *(x³-3x+11)' = (3x² -3)/ (2√(x^3-3x+11)) 

Answer 2

To find the extreme points of the function $F(x) = \sqrt{x^3} - 3x + 11$ over the interval $[-2, 13]$, we need to follow these steps:

1. Find the critical points of the function within the given interval.
2. Evaluate the function at these critical points and the endpoints of the interval.
3. Identify which points give the highest and lowest values for the function.

Step 1: Find the critical points of the function. Critical points occur when the derivative of the function is equal to zero or undefined.

First, let's find the derivative of $F(x)$: $F'(x) = \frac{d}{dx} \left( \sqrt{x^3} - 3x + 11 \right)$

Using the chain rule for differentiation, we have: $F'(x) = \frac{1}{2 \sqrt{x^3}} \cdot \frac{d}{dx} (x^3) - 3$

$F'(x) = \frac{3x^2}{2 \sqrt{x^3}} - 3$

Now, set the derivative equal to zero and solve for $x$: $\frac{3x^2}{2 \sqrt{x^3}} - 3 = 0$

To simplify, multiply both sides by $2 \sqrt{x^3}$: $3x^2 - 6 \sqrt{x^3} = 0$

Now, isolate the term containing the square root: $6 \sqrt{x^3} = 3x^2$

Square both sides to eliminate the square root: $36x^3 = 9x^4$

$9x^4 - 36x^3 = 0$

Factor out $9x^3$: $9x^3 (x - 4) = 0$

So, we have two critical points:

1. $x = 0$ (Note: This critical point is not within the interval $[-2, 13]$, so we can ignore it for this problem.)
2. $x = 4$

Step 2: Evaluate the function at the critical point and endpoints of the interval. We need to evaluate $F(x)$ at $x = 4$, $x = -2$, and $x = 13$.

$F(4) = \sqrt{4^3} - 3(4) + 11 = 8 - 12 + 11 = 7$

$F(-2) = \sqrt{(-2)^3} - 3(-2) + 11 = -2 - (-6) + 11 = 15$

$F(13) = \sqrt{13^3} - 3(13) + 11$ (Note: The value of $F(13)$ will be very large and not relevant for finding extreme points since the function increases without bound as $x$ approaches infinity.)

Step 3: Identify the extreme points. The extreme points are the highest and lowest values of the function within the interval $[-2, 13]$.

The extreme point with the lowest value is $x = 4$ (since $F(4) = 7$ is the smallest value among the critical points and endpoints).

The extreme point with the highest value is $x = -2$ (since $F(-2) = 15$ is the largest value among the critical points and endpoints).

In conclusion, the extreme points of the function $F(x) = \sqrt{x^3} - 3x + 11$