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Caxap04egg
@Caxap04egg
July 2022
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Напишите уравнение касательной к графику функции f(x)=-3x^2+4x^(1/2)+5 в точке этого графика с абсциссой x= 4
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sedinalana
Verified answer
F(4)=-48+8+5=-35
f`(x)=-6x+2/√x
f`(4)=-24+1=25
y=-35+25(x-4)=-35+25x-100=25x-135
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Verified answer
F(4)=-48+8+5=-35f`(x)=-6x+2/√x
f`(4)=-24+1=25
y=-35+25(x-4)=-35+25x-100=25x-135