Задание 1. Найти область определения функции:
x3–6x2+9x ≥ 0
x·(x2–6х+9) ≥ 0
x·(x–3)2 ≥ 0
_–__ [0] __+__ [3] __+__
О т в е т. [0;+ ∞ )
4б)
12+4x–x2>0
x2–4x–12 <0
D=(–4)2–4·(–12)=16+48=64
x1=(4–8)/2=–2; x2=(4+8)/2=6
_+__ (–2) __–__ (6) __+_
О т в е т. (–2;6)