Question

помогите пожалуйста ​

Answer 1

Объяснение:

$f(x) = \frac{ {x}^{10} }{5} - {x}^{8} + 5 \\ {f}^{l} (x) = 2 {x}^{9} - 8 {x}^{7} \\ 2 {x}^{9} - 8 {x}^{7} = 0 \\ 2 {x}^{7} ( {x}^{2} - 4) = 0 \\ x = 0 \\ x = 2 \\ x = - 2 \\ f(0) = 5 \\ f(2) = - 46.2 \\ f( - 2) = - 46.2 \\ max \: f(x) = f(0) = 5 \\ x = 0$

$f(x) = - 3 {x}^{2} + 2 {x}^{3} \\ {f}^{l} (x) = - 6x + 6 {x}^{2} \\ - 6x + 6 {x}^{2} = 0 \\ 6x(x - 1) = 0 \\ x = 0 \\ x = 1 \\ f(0) = 0 \\ f(1) = - 1 \\ min \: f(x) = f(1) = - 1 \\ x = 1$

$f(x) = x - \frac{4}{3} {x}^{3} \\ {f}^{l} (x) = 1 - 4 {x}^{2} \\ 1 - 4 {x}^{2} = 0 \\ 4 {x}^{2} = 1 \\ {x}^{2} = \frac{1}{4} \\ x = \frac{1}{2} \\ x = - \frac{1}{2} \\ f( \frac{1}{2} ) = \frac{1}{3} \\ f(0) = 0$

$f(1) = - \frac{1}{3} \\ min \: f(x) = f( 1) = - \frac{1}{3}$

Answer 2

Ответ:

4 ; 5 на фото

Объяснение:

6.

f(x)=x-4/3x³

f'(x)=(x)’-(4/3x³)'=1-4/3•3x²=1-4x²

f'(x)=0

1-4x²=0

4x²=1

x²=1/4

x= ±1/2

f(0)=0

f(1/2)=1/3

f(1)= -1/3

наименьшее значение f(x)= -1/3

ответ: В