Question

Найти наибольшее и наименьшее значение функции f(x)=x^3-2x^2-4x+2 на промежутке [-1,1]

Answer 1

f(x)=x^3-2x^2-4x+2

f'(x)=3x^2-4x-4

3x^2-4x+4=0

D=16-16*3<0

f(-1)=1-2+4+2=5 - наиб.

f(1)=1-2-4+2=-3 - наим.

Answer 2

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

Step 1: Find the critical points. Critical points occur where the derivative of the function is equal to zero or is undefined. The critical points are the potential candidates for the extreme values.

$f'(x) = 3x^2 - 4x - 4$

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

$3x^2 - 4x - 4 = 0$

Using the quadratic formula, we get:

$x = \frac{4 \pm \sqrt{4^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3}$

$x = \frac{4 \pm \sqrt{64}}{6}$

$x = \frac{4 \pm 8}{6}$

So, the critical points are $x = \frac{2}{3}$ and $x = -1$.

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 endpoints of the interval $[-1, 1]$.

$f(-1) = (-1)^3 - 2(-1)^2 - 4(-1) + 2 = -1 + 2 + 4 + 2 = 7$

$f\left(\frac{2}{3}\right) = \left(\frac{2}{3}\right)^3 - 2\left(\frac{2}{3}\right)^2 - 4\left(\frac{2}{3}\right) + 2$

$f\left(\frac{2}{3}\right) = \frac{8}{27} - 2 \cdot \frac{4}{9} - \frac{8}{3} + 2 = \frac{8 - 24 - 72 + 54}{27} = -\frac{34}{27}$

$f(1) = 1^3 - 2 \cdot 1^2 - 4 \cdot 1 + 2 = 1 - 2 - 4 + 2 = -3$

Step 3: Compare the values to find the maximum and minimum. Now, compare the values we obtained to find the maximum and minimum on the given interval.

The maximum value is 7 (attained at $x = -1$), and the minimum value is $-\frac{34}{27}$ (attained at $x = \frac{2}{3}$).

So, the maximum value of the function on the interval $[-1, 1]$ is 7, and the minimum value is $-\frac{34}{27}$.