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anna5355535
@anna5355535
July 2022
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Составь уравнение касательной к графику функции f(x)=-x^2+2 в точке х0=0
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Альфачка
F (×)=×^2+2 ×0=0
y=f'(×0)*(×-×0)+f(×0)
f'(×)=2×
f (×0)=f (0)=0^2+2=0+2=2
f'(×0)=f'(0)=2*0=0
y=0*(×-0)+2
y=0-0+2
y=2
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Answers & Comments
y=f'(×0)*(×-×0)+f(×0)
f'(×)=2×
f (×0)=f (0)=0^2+2=0+2=2
f'(×0)=f'(0)=2*0=0
y=0*(×-0)+2
y=0-0+2
y=2