Question

срочно пожалуйста ...............

Answer 1

Ответ:

а

$y = {x}^{3} - 7 {x}^{2} + 16x - 9$

$y '= 3 {x}^{2} - 14x + 16 \\ \\ y '= 0 \\ 3 {x}^{2} - 14x + 16 = 0 \\ D= 196 - 192 = 4 \\ x_1 = \frac{14 + 2}{6} = \frac{16}{6} = \frac{8}{3} \\ x_2 = \frac{12}{6} = 2$

б

$y = 2 {x}^{5} + 10x + 1$

$y' = 10 {x}^{4} + 10$

$y '= 0 \\ 10 {x}^{4} +10 = 0 \\ {x}^{4} + 1 = 0 \\ {x}^{4} = - 1$

нет критических точек