How do you find the domain of #y=4/(2(x-2)^2-1 )#?

Question

996 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-18T13:30:55

The domain is #D_y=RR-{2-1/sqrt2,2+1/sqrt2}#

#y=4/(2(x-2)^2-1)#

As we cannot divide by #0#,

#2(x-2)^2-1!=0#

#2(x-2)^2!=1#

#(x-2)^2!=1/2#

#(x-2)!=+-1/sqrt2#

#x!=2+-1/sqrt2#

The domain of #y# is #D_y=RR-{2-1/sqrt2,2+1/sqrt2}#

Answer 2 · 2017-03-18T13:33:58

#x in RR - {2-1/sqrt(2), 2+1/sqrt(2)}#

#y=4/(2(x-2)^2-1)# is defined for all values of #x# for which

#color(white)("XXX")2(x-2)^2-1 !=0#

That is for all values except
#color(white)("XXX")(x-2)^2=1/2#

#color(white)("XXX")x-2=+-1/sqrt(2)#

#color(white)("XXX")x=2+-1/sqrt(2)#