Question

F ( x ) =x⁴-2x²+3, [-2; 3 ]​

Answer 1

Ответ:

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Answer 2

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

First, let's find the critical points by taking the derivative of $f(x)$ and setting it equal to zero:

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

Setting $f'(x)$ equal to zero and solving for $x$:

$4x^3 - 4x = 0$

$4x(x^2 - 1) = 0$

This gives us critical points $x = 0$ and $x = \pm 1$.

Next, we'll evaluate $f(x)$ at these critical points and the endpoints of the interval $[-2, 3]$:

1. $f(-2)$
2. $f(0)$
3. $f(1)$
4. $f(3)$

$f(-2) = (-2)^4 - 2(-2)^2 + 3 = 16 - 8 + 3 = 11$

$f(0) = 0^4 - 2(0)^2 + 3 = 3$

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

$f(3) = 3^4 - 2(3)^2 + 3 = 81 - 18 + 3 = 66$

Now, we compare these values to find the absolute maximum and minimum on the interval $[-2, 3]$:

• Absolute maximum: $f(-2) = 11$
• Absolute minimum: $f(3) = 66$