Question

Помогите пожалуйста, 6 заданий - 100 баллов

Answer 1

Ответ:

1.

1

$f'(x) = 3 \times 6 {x}^{5} + \frac{1}{4} \times 4 {x}^{3} - 2 \times 2x + 5 = \\ = 30 {x}^{5} + {x}^{3} - 4x + 5$

2

$f'(x) = ((2 - 5x) \sqrt{x} ) '= (2 \sqrt{x} - 5x \sqrt{x} ) '= \\ = (2 {x}^{ \frac{1}{2} } - 5 {x}^{ \frac{3}{2} } ) '= 2 \times \frac{1}{2} {x}^{ - \frac{1}{2} } - 5 \times \frac{3}{2} {x}^{ \frac{1}{2} } = \frac{1}{ \sqrt{x} } - \frac{15}{2} \sqrt{x}$

3

$f'(x) = \frac{( {x}^{2} - 8x)'(x + 2) - (x + 2)'( {x}^{2} - 8x) }{ {(x + 2)}^{2} } = \\ = \frac{(2x - 8)(x + 2) - ( {x}^{2} - 8x)}{ {(x + 2)}^{2} } = \\ = \frac{2 {x}^{2} + 4x - 8x - 16 - {x}^{2} + 8x}{ {(x + 2)}^{2} } = \frac{ {x}^{2} + 4x - 16}{ {(x + 2)}^{2} }$

4

$f'(x) = (4 {x}^{ - 2} - 5 {x}^{ - 4} )' = \\ = 4 \times ( - 2) {x}^{ - 3} - 5 \times ( - 4) {x}^{ - 5} = - \frac{8}{ {x}^{3} } + \frac{20}{ {x}^{5} }$

2.

$f(x) = 3 {x}^{2} - {x}^{3} \\ x_0 = - 2$

$f(x) = f(x_0) + f'(x_0)(x - x_0)$

$f( - 2) = 3 \times 4 + 8 = 12 + 8 = 20$

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

$f'( - 2) = - 12 - 3 \times 4 = - 24$

$f(x) = 20 - 24(x + 2) = 20 - 24x - 48 = \\ = - 24x - 28$

- уравнение касательной

3.

$f'(x) = \frac{( {x}^{2} + x + 4)'(x + 1) - (x + 1)'( {x}^{2} + x + 4)(x + 1)}{ {(x + 1)}^{2} } = \\ = \frac{(2x + 1)(x + 1) - ( {x}^{2} + x + 4) }{ {(x + 1)}^{2} } = \\ = \frac{2 {x}^{2} + 2x + x + 1 - x {}^{2} - x - 4 }{ {(x + 1)}^{2} } = \\ = \frac{ {x}^{2} + 2x - 4}{ {(x + 1)}^{2} } \\ \\ \frac{ {x}^{2} + 2x - 4 }{ {(x + 2)}^{2} } = 0 \\ \\ {x}^{2} + 2x - 4 = 0 \\ D = 4 + 16 = 20 = 4 \times 5 \\ x_1 = \frac{ - 2 + 2 \sqrt{5} }{2} = - 1 + \sqrt{5} \\ x_2 = - 1 - \sqrt{5} \\ \\ x\ne - 2 \\ \\ + \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - \: \: \: \: \: \: \: \: \: \: \: - \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: + \\ - -( - 1 - \sqrt{5}) -( - 2) -( - 1 - \sqrt{5}) - - >$

(-1 - корень из 5) - точка максимума

(-1 + корень из 5) - точка минимума