Автор - ALEXMIR15

написать уравнение касательной к графику функции f(x)=3x^2, X0=-3

Ответ

Автор - 777sanyapro100

y(кас) = f'(x0) (x - x0) + f(x0)

f(x) = 3x^2

f(x0) = f(-3) = 3 × 9 = 27

f'(x) = (3x^2)' = 6x

f'(x0) = f'(-3) = -3 × 6 = -18

y(кас) = -18(x + 3) + 27 = -18x - 54 + 27 = -18x - 27

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