How do you find the domain and range of $f(x) = (2x+1) / (2x-1)$ ?

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Answer 1 · 2016-12-23T05:28:09

The domain is $x in RR-{1/2}$
The range is $f(x) in RR-{1}$

As you cannot divide by $0$ , $x!=1/2$

There is a vertical asymptote $x=1/2$

Therefore,

The domain of $f(x)$ is $D_(f(x))$ is $x in RR-{1/2}$

$lim_(x->+-oo)f(x)=lim_(x->+-oo)(2x)/2x=1$

There is a horizontal asymptote $y=1$

The range is $f(x) in RR-{1}$

graph{(y-(2x+1)/(2x-1))(y-1)(x-1/2)=0 [-12.66, 12.65, -6.33, 6.33]}

How do you find the domain and range of f(x) = (2x+1) / (2x-1)f(x) = (2x+1) / (2x-1)?