What is the domain and range of y= (2x^2-1)/(2x-1)y=2x212x1?

1 Answer
Aug 4, 2017

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

Explanation:

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]}