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vadimnoob12
@vadimnoob12
August 2022
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f(x)=1/3x^3-x^2-3x =
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багира9090
F(x)= 1/3 * x^3 + 4x^2 - 9x +1
f'(x) = x^2+8x-9f'(x)>0x^2+8x-9>0(x-1)(x+9)>0 + - +----(-9)-------(1)------->
x принадлежит (-беск; -9)U(1;+беск)
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vadimnoob12
А по какому принципу вы это решили ?
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Answers & Comments
f'(x) = x^2+8x-9f'(x)>0x^2+8x-9>0(x-1)(x+9)>0 + - +----(-9)-------(1)------->
x принадлежит (-беск; -9)U(1;+беск)