Question

Найдите область определения функции (т.е D(f) )
γ= √х²+4х-12

Answer 1

$y= sqrt{x^2+4x-12} \ \ x^2+4x-12 geq 0 \ \ x^2+4x-12=0 \ x_1=-6\ x_2=2 \ \ +++[-6]---[2]+++ textgreater x \ \ x in (-infty;-6] cup [2;+infty) \ \ OTBET: D(y)= (-infty;-6] cup [2;+infty)$

Answer 2

y=√(x²+4x-12)
x²+4x-12≥0
D=16+48=64>0
x1+x2=-4 U x1*x2=-12
x1=-6 U x2=2
             +                        _                    +
--------------------[-6]-------------------[2]--------------------
x∈(-∞;-6] U [2;∞)