Question

F(x)=x^4-x/2+1 {-1;1}найти максимальное и минимальное значение функции

Answer 1

$f(x)=x^4- \dfrac{x}{2}+2 \\ f'(x)=4x^3- \dfrac{1}{2} \\ \\ 4x^3- \dfrac{1}{2}=0 \\ 4x^3= \dfrac{1}{2} \\ x^3= \dfrac{1}{8} \\ x=0,5 \\ \\ f(-1)=(-1)^4- \dfrac{-1}{2}+1= 1+0,5+1=2,5 \\ f(0,5)=(0,5)^4- \dfrac{0,5}{2}+1=0,0625-0,25+1=0,8125 \\ f(1)=1- \dfrac{1}{2}+1=1,5$

Ответ:  $y_{min}=0,8125; \ y_{max}=2,5$

Answer 2

To find the maximum and minimum values of the function $F(x) = x^4 - \frac{x}{2} + 1$ within the interval $[-1, 1]$, we need to consider critical points and the endpoints of the interval.

1. Critical points: Critical points are the values of $x$ where the derivative of the function is equal to zero or does not exist.

First, let's find the derivative of $F(x)$ with respect to $x$: $F'(x) = 4x^3 - \frac{1}{2}$

Now, we can find the critical points by setting the derivative equal to zero and solving for $x$: $4x^3 - \frac{1}{2} = 0$ $4x^3 = \frac{1}{2}$ $x^3 = \frac{1}{8}$ $x = \sqrt[3]{\frac{1}{8}}$ $x = \frac{1}{2}$

So, the critical point within the interval $[-1, 1]$ is $x = \frac{1}{2}$.

2. Endpoints of the interval: The endpoints of the interval are $x = -1$ and $x = 1$.

Now, we need to evaluate the function at the critical point and the endpoints to find the maximum and minimum values:

1. $F(-1) = (-1)^4 - \frac{-1}{2} + 1 = 1 + \frac{1}{2} + 1 = \frac{5}{2}$
2. $F\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^4 - \frac{\frac{1}{2}}{2} + 1 = \frac{1}{16} - \frac{1}{4} + 1 = \frac{13}{16}$
3. $F(1) = 1^4 - \frac{1}{2} + 1 = 1 - \frac{1}{2} + 1 = \frac{3}{2}$

So, within the interval $[-1, 1]$, the minimum value of the function is $\frac{5}{2}$ and the maximum value is $\frac{13}{16}$.