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]}

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