Question

Y=1/3x^3-2t^2+3 Найдите наибольшее и наименьшее значение функции на отрезке (-1,2)

Answer 1

Решение на фотографии

Answer 2

To find the maximum and minimum values of the function $Y = \frac{1}{3}x^3 - 2t^2 + 3$ on the interval $(-1, 2)$, we need to analyze the critical points and endpoints of the interval.

Step 1: Find the critical points (where the derivative is either zero or undefined): Calculate the derivative of $Y$ with respect to $x$ and $t$ separately and set them equal to zero to find critical points.

$Y'_{x} = x^2$, so $x^2 = 0$ gives us a critical point at $x = 0$.

$Y'_{t} = -4t$, so $-4t = 0$ gives us a critical point at $t = 0$.

Step 2: Check the endpoints of the interval: Evaluate the function $Y$ at the endpoints of the interval $(-1, 2)$:

When $x = -1$ and $t = 0$: $Y = \frac{1}{3}(-1)^3 - 2(0)^2 + 3 = \frac{1}{3} - 2 + 3 = \frac{4}{3}$. When $x = 2$ and $t = 0$: $Y = \frac{1}{3}(2)^3 - 2(0)^2 + 3 = \frac{8}{3} + 3 = \frac{17}{3}$.

Step 3: Compare all the values obtained to find the maximum and minimum:

• The critical points: $Y(0, 0) = 3$.
• The endpoints: $Y(-1, 0) = \frac{4}{3}$ and $Y(2, 0) = \frac{17}{3}$.

So, the minimum value of the function on the interval $(-1, 2)$ is $Y_{\text{min}} = \frac{4}{3}$ and the maximum value is $Y_{\text{max}} = \frac{17}{3}$.