Question

Найти наибольшее и наименьшее значение функции: f(x)=x³-3x²+3x+2; на [-2;2]

Answer 1

f'(x)=(x³-3x²+3x+2)'=3х²-6х+3=3*(х²-2х+1)=3*(х-1)²=0; х=1∈ [-2;2]

Ищем значение функции на концах отрезка и в критической точке х=1

f(-2)=-8-12-6+2=-24 - наименьшее значение функции

f(2)=8-12+6+2=4-наибольшее значение функции

f(1)=1-3+3+2=3

Answer 2

To find the maximum and minimum values of the function $f(x) = x^3 - 3x^2 + 3x + 2$ on the interval $[-2, 2]$, we need to follow these steps:

1. Find the critical points of the function within the interval $[-2, 2]$. These are the points where the derivative of the function is equal to zero or is undefined.
2. Evaluate the function at the critical points and endpoints of the interval.
3. Compare the values obtained to identify the maximum and minimum.

Step 1: Find the critical points by taking the derivative of the function and setting it equal to zero:

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

$f'(x) = 3x^2 - 6x + 3$

Set $f'(x) = 0$ to find the critical points:

$3x^2 - 6x + 3 = 0$

Divide by 3:

$x^2 - 2x + 1 = 0$

This is a quadratic equation that can be factored as:

$(x - 1)(x - 1) = 0$

So, $x = 1$ is the only critical point within the interval $[-2, 2]$.

Step 2: Evaluate the function at the critical point and the endpoints of the interval:

For $x = -2$: $f(-2) = (-2)^3 - 3(-2)^2 + 3(-2) + 2 = -8 - 12 - 6 + 2 = -24$

For $x = 1$ (critical point): $f(1) = 1^3 - 3(1)^2 + 3(1) + 2 = 1 - 3 + 3 + 2 = 3$

For $x = 2$: $f(2) = 2^3 - 3(2)^2 + 3(2) + 2 = 8 - 12 + 6 + 2 = 4$

Step 3: Compare the values:

The minimum value of the function occurs at $x = -2$ with a value of $f(-2) = -24$.

The maximum value of the function occurs at $x = 1$ with a value of $f(1) = 3$.

To summarize:

• The minimum value of the function is -24, which occurs at $x = -2$.
• The maximum value of the function is 3, which occurs at $x = 1$.