September 2021 0 9 Report
Найди, в какой точке графика функции y=f(x) касательная параллельна заданной прямой: y=9+4x, f(x)=x^3/3−6x^2+40x−10.

Ответ (при необходимости округли с точностью до десятых):

касательная параллельна заданной прямой в точке с координатами ( ; )
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