Question

Найти найбольшее и наймешьшее значение F(x) =x4/2-9x2 на [-1;4]

Answer 1

Дано:

Y(х)= 1/2*x⁴ - 9*x²

D(y) - X∈[-1;4]

Найти: Локальные экстремумы.

Пошаговое объяснение:

Экстремумы находятся в корнях первой производной функции.

1) Y'(x) = 1/2*4*x³ - 9*2*x = 2*x³ - 18*x  = 2*x*(x²- 9) = 2*x*(x-3)*(x+3) = 0

Получили три корня: х₁ = -3 (не входит в D(y), x₂ = 0 и х₃ = 3.

Рисунок с графиком функции в приложении.

Максимум при х=0, Y(0) = 0 - ответ

Минимум при х = 3,  Y(3) =  - 40.5 - ответ

Answer 2

To find the maximum and minimum values of the function $F(x) = \frac{x^4}{2} - 9x^2$ on the interval $[-1, 4]$, we'll need to follow these steps:

1. Find the critical points of the function within the given interval.
2. Evaluate the function at the critical points and the endpoints of the interval.
3. Compare the values to determine the maximum and minimum.

Step 1: Find the critical points. Critical points occur where the derivative of the function is equal to zero or is undefined.

Let's first find the derivative of $F(x)$:

$F'(x) = \frac{d}{dx}\left(\frac{x^4}{2} - 9x^2\right) = 4 \cdot \frac{x^3}{2} - 18x$.

Now, set $F'(x) = 0$ to find the critical points:

$4 \cdot \frac{x^3}{2} - 18x = 0$.

Step 2: Solve for critical points.

$2x^3 - 18x = 0$.

Factor out $2x$:

$2x(x^2 - 9) = 0$.

Set each factor to zero:

$2x = 0$ or $x^2 - 9 = 0$.

Solving for $x$ in the second equation:

$x^2 = 9$.

$x = \pm 3$.

So, the critical points are $x = 0$ and $x = \pm 3$.

Step 3: Evaluate $F(x)$ at the critical points and the endpoints of the interval.

$F(0) = \frac{0^4}{2} - 9 \cdot 0^2 = 0$.

$F(-1) = \frac{(-1)^4}{2} - 9 \cdot (-1)^2 = \frac{1}{2} - 9 = -\frac{17}{2}$.

$F(4) = \frac{4^4}{2} - 9 \cdot 4^2 = 32 - 144 = -112$.

$F(3) = \frac{3^4}{2} - 9 \cdot 3^2 = \frac{81}{2} - 81 = -\frac{63}{2}$.

Step 4: Compare the values to find the maximum and minimum.

The maximum and minimum values occur at the critical points and endpoints.

$F(x)$ is decreasing from $-\infty$ to $-\frac{17}{2}$, then increasing from $-\frac{17}{2}$ to $-\frac{63}{2}$, and finally decreasing from $-\frac{63}{2}$ to $-112$.

The minimum value is $-\frac{63}{2}$ and occurs at $x = 3$.

The maximum value is 0 and occurs at $x = 0$ and $x = -1$.

So, the minimum value is $-\frac{63}{2}$ and the maximum value is 0.