Question

F(x)=х³-6х²+1 [-1;4] ............​

Answer 1

Ответ:

$255 \frac{2 {54 = = .2 < \times \frac{?}{?} \times \frac{?}{?} \times \frac{?}{?} }^{2} }{?}$

Answer 2

To find the minimum and maximum values of the function $f(x) = x^3 - 6x^2 + 1$ in the interval $[-1, 4]$, we need to evaluate the function at the critical points and endpoints of the interval.

1. Critical points: To find the critical points, we need to find the values of $x$ where the derivative of the function is equal to zero or does not exist.

First, let's find the derivative of $f(x)$: $f'(x) = 3x^2 - 12x$

Setting $f'(x)$ equal to zero and solving for $x$: $3x^2 - 12x = 0$ $3x(x - 4) = 0$

From this equation, we find two critical points: $x = 0$ and $x = 4$.

2. Endpoints of the interval: The interval $[-1, 4]$ has two endpoints: $x = -1$ and $x = 4$.

Now, we evaluate the function $f(x)$ at the critical points and endpoints to find the maximum and minimum values:

• Evaluate $f(x)$ at $x = -1$: $f(-1) = (-1)^3 - 6(-1)^2 + 1 = -1 - 6 + 1 = -6$

• Evaluate $f(x)$ at $x = 0$: $f(0) = (0)^3 - 6(0)^2 + 1 = 0 - 0 + 1 = 1$

• Evaluate $f(x)$ at $x = 4$: $f(4) = (4)^3 - 6(4)^2 + 1 = 64 - 96 + 1 = -31$

Therefore, the maximum value of $f(x)$ in the interval $[-1, 4]$ is 1, and the minimum value is -31.