Question

Найти наименьшее и наибольшее значение функции f(x)=x^2+8/x-1 ;[-3;0]

Answer 1

Решение задания смотри на фотографии

Answer 2

To find the minimum and maximum values of the function $f(x) = x^2 + \frac{8}{x-1}$ over the interval $[-3, 0]$, we need to analyze critical points and endpoints of the interval.

Step 1: Critical Points Critical points occur when the derivative of the function is equal to zero or undefined. Let's find the derivative of $f(x)$:

$f'(x) = \frac{d}{dx} \left(x^2 + \frac{8}{x-1}\right)$

Using the quotient rule, we get:

$f'(x) = 2x - \frac{8}{(x-1)^2}$

Now, to find critical points, we need to solve $f'(x) = 0$:

$2x - \frac{8}{(x-1)^2} = 0$

Step 1.1: Solve for $x$ when $2x - \frac{8}{(x-1)^2} = 0$:

$2x = \frac{8}{(x-1)^2}$

$(x-1)^2 = \frac{8}{2} = 4$

$x-1 = \pm \sqrt{4} = \pm 2$

Step 1.2: Find the two possible values for $x$:

1. $x - 1 = 2$: $x = 3$
2. $x - 1 = -2$: $x = -1$

So, we have two critical points: $x = 3$ and $x = -1$.

Step 2: Endpoints of the Interval The interval is $[-3, 0]$, so the endpoints are $x = -3$ and $x = 0$.

Step 3: Evaluate the Function at Critical Points and Endpoints Now, we need to evaluate the function $f(x)$ at the critical points and endpoints:

1. $x = 3$: $f(3) = 3^2 + \frac{8}{3-1} = 9 + 4 = 13$

2. $x = -1$: $f(-1) = (-1)^2 + \frac{8}{-1-1} = 1 - 4 = -3$

3. $x = -3$: $f(-3) = (-3)^2 + \frac{8}{-3-1} = 9 - 2 = 7$

4. $x = 0$: $f(0) = 0^2 + \frac{8}{0-1} = 0 - 8 = -8$

Step 4: Identify the Minimum and Maximum Values Now, we compare the function values to find the minimum and maximum values over the given interval:

• The minimum value is -8, which occurs at $x = 0$.
• The maximum value is 13, which occurs at $x = 3$.

So, the minimum value of the function $f(x) = x^2 + \frac{8}{x-1}$ over the interval $[-3, 0]$ is -8, and the maximum value is 13.