Question

Знайдіть критичні точки функції f(x)=x^2-2x^3

Answer 1

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