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Автор - russia53

пусть f (x)=x3 (x-3) решить неравенство f'(x)>0​

Ответ

Автор - 2ReCKey

f(x)=x^3(x-3)\f'(x)=3x^2(x-3)+x^3\3x^3-9x^2+x^3>0\4x^3-9x^2>0\x^2*(4x-9)>0

так как x^2 всегда положительный на него можно сократить, рассмотрим 2 случая:

x^2=0:\0>0 в данном случае решений нет, значит x≠0.

4x-9>0\4x>9\x>frac{9}{4}

Ответ: x∈(2.25;+∞)

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