Question

40 баллов. знайти точки екстремуму функції

Answer 1

$f(x) = {x}^{3} - 3 {x}^{2} + 10 \\ f'(x) = 3 {x}^{3 - 1} - 2 \times 3 {x}^{2 - 1} = 3 {x}^{2} - 6x \\ f'(x) = 0 \\ 3 {x}^{2} - 6x = 0 \\ {x}^{2} - 2x = 0 \\ x(x - 2) = 0 \\ x_{1} =0 \\ x_{2} = 2 \\ + + + [0] - - - [2] + + + \\ x_{max} = 0 \\ x_{min} = 2$