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@kkkkiiiii
March 2022
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Составьте уравнение касательной к графику функции
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unkg
F(x0)=2 f'(x)=-6/x^2 f'(x0)=-2/3
y=f(x0)+f'(x0)(x-x0)
y=2-2/3(x-3)
y=2-2/3x+2
y=4-2/3x
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Answers & Comments
y=f(x0)+f'(x0)(x-x0)
y=2-2/3(x-3)
y=2-2/3x+2
y=4-2/3x