Question

Срочно!!!!! Помогите пожалуйста!!!!

Answer 1

Ответ:

1

$f'(x) = \frac{1}{2} {( - 5 {x}^{2} + 10x + 4) }^{ - \frac{1}{2} } \times ( - 5 {x}^{2} + 10x + 4) '= \\ = \frac{1}{2 \sqrt{ - 5 {x}^{2} + 10x + 4} } \times ( - 10x + 10) = - \frac{5x + 5}{ \sqrt{ - 5 {x}^{2} + 10x + 4} }$

$f'(0) = - \frac{5}{ \sqrt{4} } = - \frac{5}{2} = - 2.5$

2

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

$f'(3) = - 6 \times \sqrt{4} + \frac{3 \times ( - 5)}{ \sqrt{4} } = \\ = - 12 - \frac{15}{2} = - 12 - 7.5 = - 19.5$

3

$f'(x) = \frac{( \sqrt{ {x}^{2} + 6x + 4} ) '\times ( {x}^{2} - 3x + 1) - ( {x}^{2} - 3x + 1) '\times \sqrt{ {x}^{2} + 6x + 4 } }{ {( {x}^{2} - 3x + 1)}^{2} } = \\ = \frac{ \frac{1}{2 \sqrt{ {x}^{2} + 6x + 4} } \times (2x + 6) \times ( {x}^{2} - 3x +1 ) - (2x - 3) \sqrt{ {x}^{2} + 6x + 4 } }{( {x}^{2} - 3x + 1) {}^{2} } = \\ = \frac{ 1}{( {x}^{2} - 3x + 1) {}^{2} } ( \frac{(x + 3)( {x}^{2} - 3x + 1)}{ \sqrt{ {x}^{2} + 6x + 4} } - (2x - 3) \sqrt{ {x}^{2} + 6x + 4} ) \\ \\ f'(0) = \frac{1}{1} ( \frac{3}{ \sqrt{4} } - ( - 3) \times 2) = \\ = 1.5 + 6 = 7.5$