What is the domain and range of # f(x)= (3x) /(x^2-1)#?

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Answer 1 · 2018-07-02T16:01:45

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

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

Therefore,

#x^2-1!=0#

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

So,

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

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

To calculate the range, let

#y=(3x)/(x^2-1)#

#=>#, #y(x^2-1)=3x#

#=>#, #yx^2-y=3x#

#=>#. #yx^2-3x-y=0#

This ia a quadratic equation in #x# and in order to have solutions, the discriminant must be #>=0#

Therefore,

#Delta=(-3)^2-4(y)(-y)>=0#

#9+4y^2>=0#

So,

#AA y in RR#, #9+4y^2>=0#

The range is #y in RR#

graph{3x/(x^2-1) [-18.02, 18.02, -9.01, 9.02]}