Question

Розв'яжіть рівняння f'(x) = 0, якщо: f(x)=-x³+3x² + 5​

Answer 1

$f(x) = { - x}^{3} + {3x}^{2} + 5 \\ f'(x) = 0 \\ f'(x) = 6x - {3x}^{2} \\ f' (x) = 0 \\ 6x - {3x}^{2} = 0 \\ 3x(2 - x) = 0 \\ 3x = 0 \\ x_1 = 0 \\ 2 - x = 0 \\ - x = - 2 \\ x_2 = 2$

Answer 2

$f(x) = - {x}^{3} + 3 {x}^{2} + 5 \\ f'(x) = - 3 {x}^{3 - 1} + 3 \times 2x {}^{2 - 1} = \\ - 3 {x}^{2} + 6x \\ \\ f'(x) = 0 \\ - 3 {x}^{2} + 6x = 0 \\ 3{x}^{2} - 6x = 0 \\ {x}^{2} - 2x = 0 \\ x(x -2 ) = 0 \\ x _{1} = 0 \\ x _{2} = 2$