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nobubi
@nobubi
August 2022
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составьте уравнение касательной к графику функции f(x)= 3x²-5x+12. x0=1
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gfirsov0071
У = f(x0) + f'(x0)(x - x0)
f'(x) = 6x - 5
f(1) = 10
f'(1) = 1
y = 10 + (x - 1) = x - 9
Ответ: у = х - 9
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Answers & Comments
f'(x) = 6x - 5
f(1) = 10
f'(1) = 1
y = 10 + (x - 1) = x - 9
Ответ: у = х - 9