Question

F(x)=4x³+1-2x на [-1;1]

Answer 1

F(x) = 4x³−2x+1 на промежутке [-1; 1]:

F(−1) = 4·(−1)³−2·(−1)+1 = 4·(−1)+2+1 = −4+3 = −1

F(1) = 4·(1)³−2·(1)+1 = 4·(1)−2+1 = 4−1 = 3

Ответ: [−1; 3].

Answer 2

To find the extreme values of the function $F(x) = 4x^3 + 1 - 2x$ on the interval $[-1, 1]$, we need to first find the critical points and then evaluate the function at these points and the endpoints of the interval.

Step 1: Find the critical points Critical points are the values of $x$ where the derivative of the function $F(x)$ is equal to zero or does not exist. So, let's find the derivative $F'(x)$:

$F'(x) = \frac{d}{dx}(4x^3 + 1 - 2x) = 12x^2 - 2.$

Now, set the derivative equal to zero and solve for $x$ to find the critical points:

$12x^2 - 2 = 0.$

Solving for $x$:

$12x^2 = 2$

$x^2 = \frac{2}{12}$

$x^2 = \frac{1}{6}$

$x = \pm \sqrt{\frac{1}{6}}$

Step 2: Evaluate the function at the critical points and endpoints

Now, we need to evaluate the function $F(x)$ at the critical points and the endpoints of the interval $[-1, 1]$.

1. $x = -1$: $F(-1) = 4(-1)^3 + 1 - 2(-1) = -4 + 1 + 2 = -1.$

2. $x = \sqrt{\frac{1}{6}}$ (approximately $0.408$): $F\left(\sqrt{\frac{1}{6}}\right) = 4\left(\sqrt{\frac{1}{6}}\right)^3 + 1 - 2\left(\sqrt{\frac{1}{6}}\right) \approx 4(0.408) + 1 - 2(0.408) \approx 1.632.$

3. $x = -\sqrt{\frac{1}{6}}$ (approximately $-0.408$): $F\left(-\sqrt{\frac{1}{6}}\right) = 4\left(-\sqrt{\frac{1}{6}}\right)^3 + 1 - 2\left(-\sqrt{\frac{1}{6}}\right) \approx 4(-0.408) + 1 - 2(-0.408) \approx -1.632.$

4. $x = 1$: $F(1) = 4(1)^3 + 1 - 2(1) = 4 + 1 - 2 = 3.$

Step 3: Determine the extreme values

The extreme values on the interval $[-1, 1]$ are the maximum and minimum values of the function among the critical points and the endpoints.

The maximum value is 3, which occurs at $x = 1$.

The minimum value is approximately -1.632, which occurs at $x \approx -\sqrt{\frac{1}{6}}$ (approximately $-0.408$).

So, the maximum value of the function is 3, and the minimum value is approximately -1.632.