Question

За все 50
Либо найти с какой книги взяты примеры!

Answer 1

Ответ:

2)

1.

$f(x) = 2 {x}^{3} + 3 {x}^{2} - 12x - 3$

$f'(x) = 6 {x}^{2} + 6x - 12$

$f'(x) = 0 \\ 6 {x}^{2} + 6x - 12 = 0 \\ {x}^{2} + x - 2 = 0 \\ D = 1 + 8 = 9 \\ x_1 = \frac{ - 1 + 3}{2} = 1 \\ x_2 = - 2$

2.

$f(x) = 2 \sqrt{x} - 5x - 45 = 2 {x}^{ \frac{1}{2} } - 5x - 45$

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

$f'(x) = 2 \\ \frac{1}{ \sqrt{x} } - 5 = 2, \: \: x > 0 \\ \frac{1}{ \sqrt{x} } = 7 \\ \sqrt{x} = \frac{1}{7} \\ x = \frac{1}{49}$

Ответ: 1)-2; 1

2)1/49

3)

1.

$f(x) = \frac{1}{5} {x}^{5} - \frac{5}{3} {x}^{3} + 6x$

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

$f(x) < 0 \\ {x}^{4} - 5 {x}^{2} + 6 < 0 \\ \\ {x}^{2} = t \\ \\ t {}^{2} - 5t + 6 < 0 \\ D = 25 - 24 = 1 \\ t_1 = 3 \\ t_2 = 2 \\ + \: \: \: \: - \: \: \: \: \: \: \: \: + \\ - -2 - - 3- - > \\ t \in(2;3) \\ \\ \left \{ {{x {}^{2} > 2 } \atop { {x}^{2} < 3} } \right. \\ \\ {x}^{2} > 2 \\ + \: \: \: \: \: \: \: \: \: \: \: \: \: \: - \: \: \: \: \: \: \: + \\ - - ( - \sqrt{2}) - - \sqrt{2} - > \\ x\in( - \infty; - \sqrt{2} )U( \sqrt{2} ; + \infty ) \\ \\ {x}^{2} < 3 \\ + \: \: \: \: \: \: \: \: \: \: - \: \: \: \: \: \: \: \: + \\ - - ( - \sqrt{3}) - - \sqrt{3} - - > \\ x\in( - \sqrt{3} ; \sqrt{3} )$

Ответ:

$x\in( - \sqrt{3} ; - \sqrt{2} )U( \sqrt{2} ; \sqrt{3} )$

2.

$f(x) = {x}^{3} + {x}^{4}$

$f'(x) = 3 {x}^{2} + 4 {x}^{3}$

$f'(x) > 0$

$3 {x}^{2} + 4 {x}^{3} > 0 \\ {x}^{2} (4x + 3) > 0 \\ - \: \: \: \: \: \: \: \: \: \: \: + \: \: \: \: \: \: + \\ - - ( - \frac{3}{4} ) - -0 - - > \\ x\in( - \frac{3}{4};0 )U(0 ;+ \infty )$

3.

$f(x) = \frac{4}{2 - 5x}$

$f'(x) = (4 {(2 - 5x)}^{ - 1} )' = \\ = - 4 {(2 - 5x)}^{ - 2} \times ( - 5) = \frac{20}{ {(2 - 5x)}^{2} }$

$f'(x) > 0$

$\frac{20}{ {(2 - 5x)}^{2} } > 0 \\ + \: \: \: \: \: + \\ - - - \frac{2}{5} - - > \\ x\in( - \infty; \frac{2}{5} )U( \frac{2}{5} ; + \infty )$