Question

F(x)=-x²/2-1 на отрезке [-1;3] найти наименьшее и наибольшее значение функции

Answer 1

f'(x)=-1/2*2*x-0

f'(x)= -x

найдем нули функции f'(x)=0

-x=0

x=0

найдем значения в концах отрезка:

f(-1)= -(-1)²/2-1= -0,5-1= -1,5

f(3)= -3²/2-1= -4,5-1= -5,5

наибольшее значение 0

наименьшее значение  -5,5

Answer 2

To find the minimum and maximum values of the function $F(x) = -\frac{x^2}{2} - 1$ on the interval $[-1, 3]$, we'll follow these steps:

1. Find critical points by finding where the derivative is equal to zero or does not exist within the interval.
2. Evaluate the function at the critical points and endpoints.
3. Compare the values obtained in step 2 to determine the minimum and maximum.

Step 1: Find the derivative of $F(x)$: $F'(x) = -\frac{d}{dx}\left(\frac{x^2}{2}\right) - \frac{d}{dx}(1)$ $F'(x) = -\frac{2x}{2}$ $F'(x) = -x$

Step 2: Find critical points: Critical points occur where the derivative is equal to zero. $-x = 0$ $x = 0$

Step 3: Evaluate the function at the critical points and endpoints: Now, we need to evaluate the function $F(x)$ at the critical points and endpoints of the interval $[-1, 3]$.

1. At $x = -1$: $F(-1) = -\frac{(-1)^2}{2} - 1 = -\frac{1}{2} - 1 = -\frac{3}{2}$

2. At $x = 0$ (this is the critical point): $F(0) = -\frac{0^2}{2} - 1 = -1$

3. At $x = 3$: $F(3) = -\frac{3^2}{2} - 1 = -\frac{9}{2} - 1 = -\frac{11}{2}$

Step 4: Compare the values: The minimum and maximum values will be the smallest and largest values among the function evaluations we found.

The minimum value: -$\frac{11}{2}$ (at $x = 3$). The maximum value: -$\frac{3}{2}$ (at $x = -1$).

So, the minimum value of the function on the interval $[-1, 3]$ is $-\frac{11}{2}$ and the maximum value is $-\frac{3}{2}$.