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

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Answer 1 · 2018-05-06T08:02:54

The domain is #x in (-oo, -1)uu(-1,1)uu(1,+oo)#. The range is #y in (-oo,-1]uu(0,+oo)#

The denominator is #!=0#

#x^2-1!=0#

#(x+1)(x-1)!=0#

#x!=-1# and #x!=1#

The domain is #x in (-oo, -1)uu(-1,1)uu(1,+oo)#

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

Therefore,

#yx^2-y=1#

#yx^2-(y+1)=0#

This is a quadratic equation in #x#

The real solutions are when the discriminant is

#Delta>=0#

#0-4*y(-(y+1))>=0#

#4y(y+1)>=0#

The solutions to this equation is obtained with a sign chart.

#y in (-oo,-1]uu(0,+oo)#

The range is #y in (-oo,-1]uu(0,+oo)#

graph{1/(x^2-1) [-7.02, 7.024, -3.51, 3.51]}