Question

Запишите уравнение касательной к графику функции f(x)=(3x-7)^3 в точке x0=3

Answer 1

f(x) = (3x - 7)^3

f(x0) = f(3) = (9 - 7)^3 = 2^3 = 8

f'(x) = ((3x - 7)^3)' = (3x - 7)' × 3(3x - 7)^2 = 3 × 3(3x - 7)^2 = 9(3x - 7)^2

f'(x0) = f(3) = 9(9 - 7)^2 = 9 × 2^2 = 9 × 4 = 36

y(кас) = f'(x0) (x - x0) + f(x0) = 36(x - 3) + 8 = 36x - 108 + 8 = 36x - 100

y(кас) = 36x - 100