What is the domain and range of #y = 1/(x^2 - 2)#?

1 Answer

Answer:

Domain: #(-oo,-sqrt2) uu (-sqrt2,sqrt2) uu(sqrt2, +oo)#
Range: #(-oo,-1/2] uu (0, +oo)#

Explanation:

The domain can be determined by equating #x^2-2=0#

solving for x:
#x=+-sqrt2# and y is not defined when #x=+-sqrt2#

therefore the domain excludes #x=+-sqrt2#

For the Range:

#y# can have a value from -1/2 and lower and values higher than zero.

So that Range: #(-oo,-1/2] uu (0, +oo)#

graph{y=1/(x^2-2)[-20,20,-10,10]}

have a nice day!