Question

Решите срочно. Математика 10 класс даю 100 балов

Answer 1

Ответ:

1

а

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

б

$f'(x) = 3 \times 5 {x}^{4} + 2 + 0 = 15 {x}^{4} + 2$

2.

а

$f'(x) = 10x - 4 \\ \\ f'(x) \geqslant 0 \\ 10x - 4 \geqslant 0 \\ x \geqslant 0.4$

б

$f'(x) = 3 - {x}^{2} \\ \\ f'(x) \geqslant 0 \\ 3 - {x}^{2} \geqslant 0 \\ ( \sqrt{3} - x)( \sqrt{3} + x) \geqslant 0 \\ - \: \: \: \: \: \: \: \: \: \: \: \: \: \: + \: \: \: \: \: \: \: \: \: \: \: \: \: \: - \\ - - ( - \sqrt{3} )- - \sqrt{3} - - > \\ x\in[- \sqrt{3} ;\sqrt{3} ]$

3.

а

$f'(x) = 10x - 11 \\ \\ 10x - 11 = - 1 \\ 10x = 10 \\ x = 1$

б

$f'(x) = - 4x + 3 \\ \\ - 4x + 3 = - 1 \\ - 4x = - 4 \\ x = 1$

4.

а

$f'(x) =( \frac{3}{2} {x}^{ - 1}) + 3 {x}^{2} - 1 = \\ = - \frac{3}{2} {x}^{ - 2} + 3 {x}^{2} - 1 = - \frac{3}{2 {x}^{2} } + 3 {x}^{2} - 1 \\ \\ f'(3) = - \frac{1}{6} + 27 - 1 = 26 - \frac{1}{6} = 25 \frac{5}{6} \\ \\ f'(1) = - \frac{3}{2} + 3 - 1 = 2 - 1.5 = 0.5$

б

$f'(x) = 2x - 1 \\ \\ f'( \frac{2}{3} ) = \frac{4}{3} - 1 = \frac{1}{3} \\ \\ f'( \frac{1}{3} ) = \frac{2}{3} - 1 = - \frac{1}{3}$

5.

$f'(x) = \frac{(2 {x + x}^{2} )'(8x + 1) - (8x + 1)'(2x + {x}^{2} )}{ {(8x + 1)}^{2} } = \\ = \frac{(2 + 2x)(8x + 1) - 8(2x + {x}^{2} )}{ {(8x + 1)}^{2} } = \\ = \frac{16x + 2 + 16 {x}^{2} + 2x - 16x - 8 {x}^{2} }{ {(8x + 1)}^{2} } = \\ = \frac{8 {x}^{2} + 2x + 2}{ {(8x + 1)}^{2} }$