Question

F(x)= x^(4) -8x^(3)-10 найти

Answer 1

Решение дано на фото.

Answer 2

To find the critical points of the function $F(x) = x^4 - 8x^3 - 10$, we need to find the values of $x$ where the derivative of the function is equal to zero.

1. Find the derivative of $F(x)$: $F'(x) = 4x^3 - 24x^2$

2. Set $F'(x)$ equal to zero and solve for $x$: $4x^3 - 24x^2 = 0$ $4x^2(x - 6) = 0$

Now, we have two possibilities for critical points:

a) $4x^2 = 0$ gives $x = 0$. b) $x - 6 = 0$ gives $x = 6$.

3. Now, we need to check if these points are indeed critical points. To do that, we find the second derivative $F''(x)$ and evaluate it at each critical point.

$F''(x) = 12x^2 - 48x$

For $x = 0$: $F''(0) = 12(0)^2 - 48(0) = 0$

For $x = 6$: $F''(6) = 12(6)^2 - 48(6) = 12(36) - 288 = 432 - 288 = 144$

4. Analyze the results:

a) For $x = 0$, since $F''(0) = 0$, the second derivative test is inconclusive. We need to perform further analysis.

b) For $x = 6$, since $F''(6) = 144$ (which is positive), the second derivative test tells us that the function has a local minimum at $x = 6$.

5. Determine the type of critical point:

a) For $x = 0$, we can't determine whether it's a maximum, minimum, or an inflection point based on the second derivative test. We may use other methods like the first derivative test or further analysis to determine its nature.

b) For $x = 6$, we know it's a local minimum.

So, the critical points of the function $F(x) = x^4 - 8x^3 - 10$ are:

• $x = 0$ (undetermined nature)
• $x = 6$ (local minimum)