Предмет: Алгебра, автор: vvsdfvs

найдите экстремумы функции f(x)=x^3-2x^2+x+3

Ответы

Автор ответа: Аноним

0

Найдем производную функции:
$f'(x)=(x^3-2x^2+x+3)'=3x^2-4x+1$
Найдем критические точки:
$3x^2-4x+1=0\ D=16-12=2\ x_1=1\ x_2=1/3$

___+__(1/3)__-___(1)___+___
Производная функции в точке х=1/3 меняет знак с (+) на (-), следовательно, х=1/3 - точка максимума, х=1 - точка минимума.

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