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llasmos
@llasmos
July 2022
1
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найдите промежутки возрастания функции:
[tex]f(x) = x ^{3} - 6x^{2} - 9[/tex]
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Armenia2780
F(x)=x³-6x²-9;x€R
1)f'(x)=3x²-12x
2)3x²-12x>0
3)3x²-12x<0
3x(x-4)>0
__+___0__-____4__+__
x€(-oo;0)U(4;+oo) функция возрастает
x€(0;4) функция убывает
1 votes
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Answers & Comments
1)f'(x)=3x²-12x
2)3x²-12x>0
3)3x²-12x<0
3x(x-4)>0
__+___0__-____4__+__
x€(-oo;0)U(4;+oo) функция возрастает
x€(0;4) функция убывает