Объяснение:
1)
[tex]f(x)=x^3-3x.\\\\f'(x)=(x^3+3x)'=3x^2-3=3*(x^2-1)=3*(x+1)*(x-1).\\\\3*(x+1)*(x-1) < 0\ |:3\\\\(x+1)*(x-1) < 0.[/tex]
-∞__+__-1__-__1__+__+∞ ⇒
Ответ: x∈(-1;1).
2)
[tex]f(x)=x^2-x^3.\\\\f'(x)=(x^2-x^3)'=2x-3x^2.\\\\2x-3x^2 < 0\\\\x*(2-3x) < 0\\\\[/tex]
-∞__-__0__+__2/3__-__+∞ ⇒
Ответ: x∈(-∞;0)U(2/3;+∞).
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Объяснение:
1)
[tex]f(x)=x^3-3x.\\\\f'(x)=(x^3+3x)'=3x^2-3=3*(x^2-1)=3*(x+1)*(x-1).\\\\3*(x+1)*(x-1) < 0\ |:3\\\\(x+1)*(x-1) < 0.[/tex]
-∞__+__-1__-__1__+__+∞ ⇒
Ответ: x∈(-1;1).
2)
[tex]f(x)=x^2-x^3.\\\\f'(x)=(x^2-x^3)'=2x-3x^2.\\\\2x-3x^2 < 0\\\\x*(2-3x) < 0\\\\[/tex]
-∞__-__0__+__2/3__-__+∞ ⇒
Ответ: x∈(-∞;0)U(2/3;+∞).