Категория:Задачник Кузнецова Графики Задача 5

В этой категории варианты Задачи 5, раздела "Графики" в задачнике Кузнецова Л.А.

Варианты[править]

Исследовать поведение функций в окрестностях заданных точек с помощью производных высших порядков.

Задача	Условие
5-1	${\displaystyle y=x^{2}-4x-(x-2)\cdot \ln {(x-1)}}$	${\displaystyle x_{0}=2}$
5-2	${\displaystyle y=4x-x^{2}-2\cos {(x-2)}}$	${\displaystyle x_{0}=2}$
5-3	${\displaystyle y=6e^{x-2}-x^{3}+3x^{2}-6x}$	${\displaystyle x_{0}=2}$
5-4	${\displaystyle y=2\ln {(x+1)}-2x+x^{2}+1}$	${\displaystyle x_{0}=0}$
5-5	${\displaystyle y=2x-x^{2}-2\cos {(x-1)}}$	${\displaystyle x_{0}=1}$
5-6	${\displaystyle y=\cos ^{2}{(x+1)}+x^{2}+2x}$	${\displaystyle x_{0}=-1}$
5-7	${\displaystyle y=2\ln {x}+x^{2}-4x+3}$	${\displaystyle x_{0}=1}$
5-8	${\displaystyle y=1-2x-x^{2}-2\cos {(x+1)}}$	${\displaystyle x_{0}=-1}$
5-9	${\displaystyle y=x^{2}+6x+8-2e^{x+2}}$	${\displaystyle x_{0}=-2}$
5-10	${\displaystyle y=4x+x^{2}-2e^{x+1}}$	${\displaystyle x_{0}=-1}$
5-11	${\displaystyle y=(x+1)\cdot \sin {(x+1)}-2x-x^{2}}$	${\displaystyle x_{0}=-1}$
5-12	${\displaystyle y=6e^{x-1}-3x-x^{3}}$	${\displaystyle x_{0}=1}$
5-13	${\displaystyle y=2x+x^{2}-(x+1)\cdot \ln {(2+x)}}$	${\displaystyle x_{0}=-1}$
5-14	${\displaystyle y=\sin ^{2}{(x+1)}-2x-x^{2}}$	${\displaystyle x_{0}=-1}$
5-15	${\displaystyle y=x^{2}+4x+\cos ^{2}{(x+2)}}$	${\displaystyle x_{0}=-2}$
5-16	${\displaystyle y=x^{2}+2\ln {(x+2)}}$	${\displaystyle x_{0}=-1}$
5-17	${\displaystyle y=4x-x^{2}+(x-2)\cdot \sin {(x-2)}}$	${\displaystyle x_{0}=2}$
5-18	${\displaystyle y=6e^{x}-x^{3}-3x^{2}-6x-5}$	${\displaystyle x_{0}=0}$
5-19	${\displaystyle y=x^{2}-2x-2e^{x-2}}$	${\displaystyle x_{0}=2}$
5-20	${\displaystyle y=\sin ^{2}{(x+2)}-x^{2}-4x-4}$	${\displaystyle x_{0}=-2}$
5-21	${\displaystyle y=\cos ^{2}{(x-1)}+x^{2}-2x}$	${\displaystyle x_{0}=1}$
5-22	${\displaystyle y=x^{2}-2x-(x-1)\cdot \ln {x}}$	${\displaystyle x_{0}=1}$
5-23	${\displaystyle y=(x-1)\cdot \sin {(x-1)}+2x-x^{2}}$	${\displaystyle x_{0}=1}$
5-24	${\displaystyle y=x^{2}-4x+\cos ^{2}{(x-2)}}$	${\displaystyle x_{0}=2}$
5-25	${\displaystyle y=x^{4}+4x^{3}+12x^{2}+24\left(x+1-e^{x}\right)}$	${\displaystyle x_{0}=0}$
5-26	${\displaystyle y=\sin ^{2}{(x-2)}-x^{2}+4x-4}$	${\displaystyle x_{0}=2}$
5-27	${\displaystyle y=6e^{x+1}-x^{3}-6x^{2}-15x-16}$	${\displaystyle x_{0}=-1}$
5-28	${\displaystyle y=\sin {x}+\operatorname {sh} {x}-2x}$	${\displaystyle x_{0}=0}$
5-29	${\displaystyle y=\sin ^{2}{(x-1)}-x^{2}+2x}$	${\displaystyle x_{0}=1}$
5-30	${\displaystyle y=\cos {x}+\operatorname {ch} {x}}$	${\displaystyle x_{0}=0}$
5-31	${\displaystyle y=x^{2}-2e^{x-1}}$	${\displaystyle x_{0}=1}$