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superdmitrui
@superdmitrui
August 2022
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Составить уравнение касательной к данной кривой в точке с абсциссой Χ₀
Υ=Χ/(Χ²+1), Χ₀= -2
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Answers & Comments
jeremy10000
Y'=f'(x)
1)f'(x0)=[(x')·(x^2+1)-(x^2+1)'·x]/(x^2+1)^2
f'(x0)=[(x^2+1)-2x^2]/(x^2+1)^2
f'(x0)=[(4+1)-8]/(4+1)^2
f'(x0)=-3/25=-0,12
2)f(x0)=-2/(4+1)
f(x0)=-2/5=-0,4
y-уравнение касательной
y=f'(x0)(x-x0)+f(x0)
y=-0,12x+0,24-0,4=-0,12x -0,16
2 votes
Thanks 1
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Answers & Comments
1)f'(x0)=[(x')·(x^2+1)-(x^2+1)'·x]/(x^2+1)^2
f'(x0)=[(x^2+1)-2x^2]/(x^2+1)^2
f'(x0)=[(4+1)-8]/(4+1)^2
f'(x0)=-3/25=-0,12
2)f(x0)=-2/(4+1)
f(x0)=-2/5=-0,4
y-уравнение касательной
y=f'(x0)(x-x0)+f(x0)
y=-0,12x+0,24-0,4=-0,12x -0,16