Question

Найдите значение производной второго порядка для функции.Помогите пожалуйста.​

Answer 1

Ответ:

1.

$f (x)= {x}^{ 3} - 3 {x}^{2} - 1 \\ f'(x)= 3 {x}^{2} - 6 \\ f''(x) = 6x \\ \\ f''( - 1) = - 6$

2.

$f (x)= {x}^{4} - {x}^{3} - x \\ f'(x) = 4 {x}^{3} - 3 {x}^{2} - 1 \\ f''(x)= 12 {x}^{2} - 6x \\ \\ f''(2) = 12 \times 4 - 12 = 12 \times 3 = 36$

3.

$f(x) = \sqrt{3 - x} \\ f'(x) = \frac{1}{2 \sqrt{3 - x} } \times ( - 1) = - 2 {(3 - x)}^{ - \frac{1}{2} }$

$f''(x) = - 2 \times ( - \frac{1}{2} ) \times {(3 - x)}^{ - \frac{3}{2} } \times ( - 1) = \\ = - \frac{1}{ \sqrt{ {(3 - x)}^{3} } }$

$f'( - 1) = - \frac{1}{ \sqrt{ {4}^{3}} } = - \frac{1}{8}$

4.

$f(x) = \sqrt{2x + 1}$

$f'(x) = \frac{1}{2 \sqrt{2x + 1} } \times 2 = \frac{1}{ \sqrt{2x + 1} }$

$f''(x) = - \frac{1}{2} {(2x + 1)}^{ - \frac{3}{2} } \times 2 = - \frac{1}{ \sqrt{ {(2x + 1)}^{3} } }$

$f''(4) = - \frac{1}{ \sqrt{ {9}^{3}} } = \frac{1}{27}$