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

Question

967 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-06T11:41:22

Domain: #x|oo,-sqrt(2))uu(-sqrt(2),sqrt(2))uu(sqrt(2),oo)#
Range: #y| (-oo,0)uu(0,oo)#

#y=1/(x^2-2) = 1/((x+sqrt2)(x-sqrt2))# . Function is undefined

if denominator is zero.So function is undefined at #x=sqrt2#

and at #x=-sqrt2#. Domain : Any real number of #x# except

#x=+-sqrt2#. Domain: #x in RR | x !=+-sqrt2#. In interval notation

expressed as #x|(-oo , -sqrt(2))uu(-sqrt(2),sqrt(2))uu(sqrt(2),oo)#.

Range: Any real number of #y# except #y=0#

Range: #y in RR | y !=0#.In interval notation expressed as

#y| (-oo,0)uu(0,oo)#

graph{1/(x^2-2) [-10, 10, -5, 5]} [Ans]