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misskholomeenkova
@misskholomeenkova
June 2022
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НАЙТИ УГЛОВОЙ КОЭФФИЦИЕНТ КАСАТЕЛЬНОЙ К ГРАФИКУ ФУНКЦИИ f(x) =4x^2 +12x+3 в точке х0=-2
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kirichekov
Verified answer
k=f'(x₀)
f(x)=4x²+12x+3, x₀=-2
f'(x)=(4x²+12x+3)'=4*2x+12=8x+12
f'(x₀)=f'(-2)=-8*(2)+12=-16+12=-4
k=-4
2 votes
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misskholomeenkova
спасибо огромное:*
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Answers & Comments
Verified answer
k=f'(x₀)f(x)=4x²+12x+3, x₀=-2
f'(x)=(4x²+12x+3)'=4*2x+12=8x+12
f'(x₀)=f'(-2)=-8*(2)+12=-16+12=-4
k=-4