odpoveď:
#S: x in -oo; 0 uu 1 + sqrt2; + oo #
vysvetlenie:
# 1 / x <= | x-2 | #
#D_f: xv RR ^ "*" #
pre #X <0 #:
# 1 / x <= - (X-2) #
# 1> -x²-2x #
# X² + 2x + 1> 0 #
# (X + 1) ²> 0 #
#x v RR ^ "*" #
Ale tu máme podmienku #X <0 #, takže:
# S_1: xv RR _ "-" ^ "*" #
Teraz, ak #X> 0 #:
# 1 / x <= X-2 #
# 1 <= x²-2x #
# X²-2x-1> = 0 #
#Δ=8#
# X 1 = (2 + sqrt8) / 2 = 1 + sqrt2 #
#cancel (x_2 = 1-sqrt2) # (#<0#)
tak # S_2: xv 1 + sqrt2; + oo #
konečne # S = S_1uuS_2 #
#S: x in -oo; 0 uu 1 + sqrt2; + oo #
0 / tu je naša odpoveď!
