Question

Y=-x^2+8x+5, [2;3] наибольшее и наименьшее значения функции

Answer 1

Ответ:

$y_{min}=17,$ при x=2; $y_{max}=20,$ при x=3

Пошаговое объяснение:

$y(2)=-2^2+8*2+5=-4+16+5=12+5=17\\y(3)=-3^2+8*3+5=-9+24+5=15+5=20$

Answer 2

To find the maximum and minimum values of the function $y = -x^2 + 8x + 5$ over the interval $[2, 3]$, we need to first analyze the critical points within the interval and then evaluate the function at the endpoints.

1. Critical Points: Critical points occur where the derivative of the function is equal to zero or does not exist. Let's find the derivative of $y$ with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(-x^2 + 8x + 5)$

$\frac{dy}{dx} = -2x + 8$

To find critical points, we set $\frac{dy}{dx}$ equal to zero and solve for $x$:

$-2x + 8 = 0$ $-2x = -8$ $x = 4$

2. Interval Endpoints: We need to evaluate the function at the endpoints of the interval $[2, 3]$, which are $x = 2$ and $x = 3$.

Now, we can find the values of $y$ at the critical point and interval endpoints:

For $x = 2$: $y = -2^2 + 8(2) + 5 = -4 + 16 + 5 = 17$

For $x = 3$: $y = -3^2 + 8(3) + 5 = -9 + 24 + 5 = 20$

For $x = 4$ (the critical point): $y = -4^2 + 8(4) + 5 = -16 + 32 + 5 = 21$

Now we can compare these values to determine the maximum and minimum within the given interval:

The maximum value is $y = 21$ (occurs at $x = 4$).

The minimum value is $y = 17$ (occurs at $x = 2$).

So, the function has a maximum value of 21 and a minimum value of 17 over the interval $[2, 3]$.