x^2-3x+2≥0,
x^2-3x+2=0,
x1=1, x2=2,
(x-1)(x-2)≥0,
x∈(-∞;1]U[2;+∞),
f'(x)=1/(2√(x^2-3x+2)) ·(x^2-3x+2)'=(2x-3)/(2√(x^2-3x+2)).
x^2-6x+9≥0
x^2-6x+9=0,
(x-3)^2=0,
(x-3)^2≥0,
x∈R,
f'(x)=3/(2√(x^2-6x+9)) ·(x^2-6x+9)'=3(2x-6)/(2√(x^2-6x+9))=(3x-9)/(√(x^2-6x+9))