Question

очень нужна помощь ​

Answer 1

Ответ:

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

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

$f'( - 1) = 3 {e}^{3 \times ( - 1) + 3} + 6 \times {( - 1)}^{2} = \\ 3 {e}^{ - 3 + 3} + 6 \times 1 = 3 {e}^{0} + 6 = 3 \times 1 + 6 = 9$