-x^2 + 6x - 5 = 0
ymax = -D/4a = 2, vetvi napravleni vniz.
Togda sqrt{ -x^{2} + 6x - 5} < 3, D(x) prinadlezhit [x1;x2] = [1;5]
togda modul otr vir = 3- sqrt{-x^{2}+6x-5}
f(x) = 3 + x^3+6x^2, D(x) = [1;5]
F'(x) = 3x^2 + 12x
F'(x) = 0 => x(x+4) = 0, x=0, x=-4
F''(x) = 6x+12
F''(0) = 12 > 0 => f(x) vozrastaet pri vseh x prinadlezhashih ODZ
xmin = 1
