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solovj
@solovj
August 2022
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Найдите точку максимума функции f(x)=x^3+x^2-8x
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Support001
Y=x^3+x^2-8x
y'=3x^2+2x-8
D=4-4*3*-8=100
x_{1,2} = -2+/-10/6
x_1 = 8/6 ; x_2 = -2
(x-8/6)(x+2)=0
_____-2_______8/6______
+ - +
-2 - точка максимума.
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Answers & Comments
y'=3x^2+2x-8
D=4-4*3*-8=100
x_{1,2} = -2+/-10/6
x_1 = 8/6 ; x_2 = -2
(x-8/6)(x+2)=0
_____-2_______8/6______
+ - +
-2 - точка максимума.