Предмет: Алгебра, автор: danikama2

СРОЧНО ДАЮ 60 БАЛЛОВ!!!!!

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Автор ответа: Аноним
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Ответ:во вложении смотрите

Объяснение: там же

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Автор ответа: NNNLLL54
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1)\; \; f(x)=x^4-8x^2-3\; \; ,\; \; x\in [-3;1\; ]\\\\f'(x)=4x^3-16x=4x\, (x^2-4)=4x\, (x-2)(x+2)=0\\\\x_1=0\; ,\; x_2=-2\; ,\; x_3=2\notin [-3;1\; ]\\\\f(-3)=(-3)^4-8\cdot (-3)^2-3=81-72-3=6\\\\f(-2)=16-32-3=-19\\\\f(0)=-3\\\\f(1)=1-8-3=-10\\\\f_{naimen.}=f(-2)=-19\; \; ,\; \; f_{naibol.}=f(-3)6

2)\; \; f(x)=x^4-4x^3+1\; \; ,\; \; x\in [-1;\, 1\; ]\\\\f'(x)=4x^3-12x^2=4x^2\, (x-3)=0\; \; ,\\\\x_1=0\; ,\; x_2=3\notin [-1;\, 1\; ]\\\\f(-1)=1+4+1=6\\\\f(0)=1\\\\f(1)=1-4+1=-2\\\\f_{naimen.}=f(1)=-2\;\ ; ,\; \; f_{naibol.}=f(-1)=6

3)\; \; x=2t^3-6t\; \; \; ,\; \; \; t\in [\; 0;2\; ]\\\\x'=6t^2-6=6(t-1)(t+1)=0\; \; \to \; \; t_1=-1\notin [\, 0;2\, ]\; \; ,\; \; t_2=1\\\\x(1)=2-6=-4\\\\x(0)=0\\\\x(2)=16-12=4\\\\x_{naimen.}=x(1)=-4\; \; ,\; \; \; x_{naibol.}=x(2)=4

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