Question

найдите точки экстремума функции
f(x)=2x^3-3x^2-1

Answer 1

$f'(x) = 6x^2 - 6x\\f'(x) = 0\\6x^2 - 6x = 0;;|:6\\x(x-1) = 0\\x = 0 ;;;;; x = 1\\+++(0)---(1)+++\\x= 0-max\\x = 1 - min ;;$