Question

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

Answer 1

Решение на фотографии, думаю на отрезке [-2,2] так должно быть

Answer 2

To find the maximum and minimum values of the function $Y = \frac{1}{3}x^3 + \frac{1}{2}x^2 - 2x - \frac{1}{3}$ on the interval $(-2, 2)$, we need to follow these steps:

1. Find the critical points by setting the derivative equal to zero and solving for $x$.
2. Determine if the critical points lie within the interval $(-2, 2)$.
3. Evaluate the function at the critical points and the endpoints of the interval.
4. Compare all these values to find the maximum and minimum.

Step 1: Find the derivative of $Y$ with respect to $x$:

$Y' = \frac{d}{dx}\left(\frac{1}{3}x^3 + \frac{1}{2}x^2 - 2x - \frac{1}{3}\right)$ $Y' = x^2 + x - 2$

Step 2: Find the critical points by setting $Y'$ equal to zero and solving for $x$:

$x^2 + x - 2 = 0$

We can solve this quadratic equation by factoring or using the quadratic formula. Factoring, we have:

$(x + 2)(x - 1) = 0$

Setting each factor to zero and solving for $x$, we get:

$x + 2 = 0 \Rightarrow x = -2$ $x - 1 = 0 \Rightarrow x = 1$

Step 3: Determine if the critical points lie within the interval $(-2, 2)$. Since both critical points $x = -2$ and $x = 1$ are within the interval $(-2, 2)$, we need to evaluate the function at these critical points and the endpoints of the interval:

$Y(-2) = \frac{1}{3}(-2)^3 + \frac{1}{2}(-2)^2 - 2(-2) - \frac{1}{3} = \frac{4}{3}$ $Y(1) = \frac{1}{3}(1)^3 + \frac{1}{2}(1)^2 - 2(1) - \frac{1}{3} = -\frac{5}{6}$ $Y(-2) = \frac{1}{3}(2)^3 + \frac{1}{2}(2)^2 - 2(2) - \frac{1}{3} = -\frac{4}{3}$

Step 4: Compare the values to find the maximum and minimum. The maximum value occurs at $x = -2$ and the minimum value occurs at $x = 1$.

So, the maximum value on the interval $(-2, 2)$ is $Y_{\text{max}} = \frac{4}{3}$, and the minimum value is $Y_{\text{min}} = -\frac{5}{6}$.