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

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Answer 1 · 2017-08-04T06:59:27

The domain is #D_f(x)=RR-{1/2}#
The range is #y in RR#

Our function is

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

The denominator cannot be #=0#

So, #2x-1!=0#, #x!=1/2#

Therefore,

The domain of #f(x)# is #D_f(x)=RR-{1/2}#

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

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

#2x^2-1=2yx-y#

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

In order for this quadratic equation in #x^2# to have solutions, the discriminant is #>=0#

#Delta=b^2-4ac=(-2y)^2-4*(2)*(y-1)>=0#

#4y^2-8(y-1)>=0#

#y^2-2y+1>=0#

#(y-1)^2>=0#

#AA y in RR#, #(y-1)^2>=0#

The range is #y in RR#

graph{(2x^2-1)/(2x-1) [-8.89, 8.89, -4.444, 4.445]}