решите уравнение f"(x)=0, если f(x)=2/3 x^3+1/2 x^2-3x
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f(x)=(2/3)*x³+(1/2)*x²-3xf ''(=[(2/3)*x³+(1/2)*x²-3x)] ''= [ (2/3)*3x²+(1/2)*2x-3] '=(2x²+x-3)' =4x+1