Категория:Задачник Кузнецова Графики Задача 1

В этой категории варианты Задачи 1, раздела "Графики" в задачнике Кузнецова Л.А.

Варианты[править]

Построить графики функций с помощью производной первого порядка.

Задача	Условие	Задача	Условие
1-1	${\displaystyle y=2x^{3}-9x^{2}+12x-9}$	1-2	${\displaystyle y=3x-x^{3}}$
1-3	${\displaystyle y=x^{2}\cdot (x-2)^{2}}$	1-4	${\displaystyle y={\frac {x^{3}-9x^{2}}{4}}+6x-9}$
1-5	${\displaystyle y=2-3x^{2}-x^{3}}$	1-6	${\displaystyle y=(x+1)^{2}\cdot (x-1)^{2}}$
1-7	${\displaystyle y=2x^{3}-3x^{2}-4}$	1-8	${\displaystyle y=3x^{2}-2-x^{3}}$
1-9	${\displaystyle y=(x-1)^{2}\cdot (x-3)^{2}}$	1-10	${\displaystyle y={\frac {x^{3}+3x^{2}}{4}}-5}$
1-11	${\displaystyle y=6x-8x^{3}}$	1-12	${\displaystyle y=16x^{2}\left({x-1}\right)^{2}}$
1-13	${\displaystyle y=2x^{3}+3x^{2}-5}$	1-14	${\displaystyle y=2-12x^{2}-8x^{3}}$
1-15	${\displaystyle y=(2x+1)^{2}\cdot (2x-1)^{2}}$	1-16	${\displaystyle y=2x^{3}+9x^{2}+12x}$
1-17	${\displaystyle y=12x^{2}-8x^{3}-2}$	1-18	${\displaystyle y=(2x-1)^{2}\cdot (2x-3)^{2}}$
1-19	${\displaystyle y={\frac {27(x^{3}-x^{2})}{4}}-4}$	1-20	${\displaystyle y={\frac {x(12-x^{2})}{8}}}$
1-21	${\displaystyle y={\frac {x^{2}(x-4)^{2}}{16}}}$	1-22	${\displaystyle y={\frac {27(x^{3}+x^{2})}{4}}-5}$
1-23	${\displaystyle y={\frac {16-6x^{2}-x^{3}}{8}}}$	1-24	${\displaystyle y=-{\frac {(x^{2}-4)^{2}}{16}}}$
1-25	${\displaystyle y=16x^{3}-36x^{2}+24x-9}$	1-26	${\displaystyle y={\frac {6x^{2}-x^{3}-16}{8}}}$
1-27	${\displaystyle y=-{\frac {(x-2)^{2}\cdot (x-6)^{2}}{16}}}$	1-28	${\displaystyle y=16x^{3}-12x^{2}-4}$
1-29	${\displaystyle y={\frac {11+9x-3x^{2}-x^{3}}{8}}}$	1-30	${\displaystyle y=-{\frac {(x+1)^{2}\cdot (x-3)^{2}}{16}}}$
1-31	${\displaystyle y=16x^{3}+12x^{2}-5}$