Question

F(x)=2x^2 -8/x+3 ,[-5 ;1] найдите наименьшее и наибольшее значения функции

Answer 1

f'=4x+8/x^2

f'(x)=0

4x^3+8=0   x^3=-2  x=-2^(1/3)

f''=4-16x^(-3)  f''(-2^(1/3))=4-16*(-2^(1/3))^(-3)=

=4-16*(-2)^(-1)>0 точка минимума

f(1)=2-8/1+3=2-8+3=-3 точка минимума

f(-5)=2*25+8/5+3=54,6  максимум

f(-2^(1/3))=2*2^(2/3)-8/(-2^(1/3))+3>0

Answer 2

To find the minimum and maximum values of the function $F(x) = 2x^2 - \frac{8}{x} + 3$ over the interval $[-5, 1]$, we need to evaluate the function at the critical points and endpoints of the interval.

Step 1: Find the 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 $F(x)$:

$F'(x) = \frac{d}{dx}(2x^2) - \frac{d}{dx}\left(\frac{8}{x}\right) + \frac{d}{dx}(3)$

$F'(x) = 4x + \frac{8}{x^2}$

Now, let's find where the derivative is equal to zero:

$4x + \frac{8}{x^2} = 0$

To solve this equation, we need to clear the fraction:

$4x^3 + 8 = 0$

$4x^3 = -8$

$x^3 = -2$

$x = \sqrt[3]{-2}$

The real cube root of -2 is approximately $x \approx -1.26$.

Step 2: Evaluate the function at the critical points and endpoints.

The critical point $x \approx -1.26$ lies within the interval $[-5, 1]$. Let's evaluate the function at the critical point and endpoints:

$F(-5) = 2(-5)^2 - \frac{8}{-5} + 3 \approx 50 + 1.6 + 3 = 54.6$

$F(-1.26) = 2(-1.26)^2 - \frac{8}{-1.26} + 3 \approx 3.18 + 6.34 + 3 = 12.52$

$F(1) = 2(1)^2 - \frac{8}{1} + 3 = 2 - 8 + 3 = -3$

Step 3: Determine the minimum and maximum values.

The minimum value of the function occurs at $x \approx -1.26$ and is approximately $F(-1.26) \approx 12.52$.

The maximum value of the function occurs at one of the endpoints, either at $x = -5$ or $x = 1$. The maximum value is approximately $F(-5) \approx 54.6$ since it is greater than $F(1) \approx -3$.

Therefore, the minimum value of the function over the interval $[-5, 1]$ is approximately 12.52, and the maximum value is approximately 54.6.