Antwoord:
Het domein is #x in -oo, 2 uu 3, + oo #
Uitleg:
#f (x) = (x-1) / (2-x) #
#G (x) = sqrt (x + 2) #
# (GOF) (x) = g (f (x)) #
# = G ((x-1) / (2-x)) #
# = Sqrt ((x-1) / (2-x) 2) #
# = Sqrt (((x-1) 2 (2-x)) / (2-x)) #
# = Sqrt ((x-1 + 4-2x) / (2-x)) #
# = Sqrt ((3-x) / (2-x)) #
daarom
# (3-x) / (2-x)> = 0 # en #x! = 0 #
Om deze ongelijkheid op te lossen, doen we een tekenkaart
#color (wit) (aaaa) ##X##color (wit) (AAAAA) ## -Oo ##color (wit) (aaaaaa) ##2##color (wit) (AAAAAAA) ##3##color (wit) (aaaaaa) ## + Oo #
#color (wit) (aaaa) ## 2-x ##color (wit) (AAAAA) ##+##color (wit) (aaa) ## ##color (wit) (aaa) ##-##color (wit) (AAAAA) ##-#
#color (wit) (aaaa) ## 3 x ##color (wit) (AAAAA) ##+##color (wit) (aaa) ## ##color (wit) (aaa) ##+##color (wit) (AAAAA) ##-#
#color (wit) (aaaa) ##G (f (x)) ##color (wit) (aaaa) ##+##color (wit) (aaa) ## ##color (wit) (aaa) ##O/##color (wit) (aaaaaa) ##+#
daarom
#G (f (x)> = 0) #, wanneer #x in -oo, 2 uu 3, + oo #
Het domein is #D_g (f (x)) # is #x in -oo, 2 uu 3, + oo #
