What is the domain and range of #y= 1/(x^2-25)#?

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Answer 1 · 2018-03-18T08:13:51

The domain of #y# is # x in RR-{-5,5}#.
The range is #y in [-1/25, 0)uu(0, +oo)#

As you cannot divide by #0#, the denominator is #!=0#

Therefore,

#x^2-25!=0#, #=># #x!=-5# and #x!=5#

The domain of #y# is #x in RR-{-5,5}#

To calculate the range, proceed as follows

#y=1/(x^2-25)#

#y(x^2-25)=1#

#yx^2-1-25y=0#

#x^2=(1+25y)/y#

#x=sqrt((1+25y)/y)#

Therefore,

#y !=0#

and

#1+25y>=0#

#y>=-1/25#

The range is #y in [-1/25, 0)uu(0, +oo)#

graph{1/(x^2-25) [-6.24, 6.244, -3.12, 3.12]}