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Kaktys03
@Kaktys03
July 2022
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уравнение касательной к графику функции f(x)=1-3x/x^2+1 в точке с абциссой х0=0
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AnonimusPro
Verified answer
Уравнение касательной:
y=f(x0)+f'(x0)*(x-x0)
x0=0
f(x0)=f(0)=1/1=1
f'(x0)=((-3)*(x^2+1)-2x*(1-3x))/(x^2+1)^2=(3x^2-2x-3)/(x^2+1)^2=f(0)=-3/1=-3
y=1-3(x-0)=1-3x
Ответ: y=1-3x
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Answers & Comments
Verified answer
Уравнение касательной:y=f(x0)+f'(x0)*(x-x0)
x0=0
f(x0)=f(0)=1/1=1
f'(x0)=((-3)*(x^2+1)-2x*(1-3x))/(x^2+1)^2=(3x^2-2x-3)/(x^2+1)^2=f(0)=-3/1=-3
y=1-3(x-0)=1-3x
Ответ: y=1-3x