What is the range of the function $f(x)=2/(3x-1)$ ?

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Answer 1 · 2017-03-19T06:30:20

The range of $f(x)$ is $R_(f(x))=RR-{0}$

Let $y=2/(3x-1)$

Then,

$(3x-1)=2/y$

$3x=2/y+1=(2+y)/y$

$x=(2+y)/(3y)$

The inverse function of $f(x)$ is

$f^-1(x)=(2+x)/(3x)$

The range of $f(x)$ is $=$ the domain of $f^-1(x)$

As we cannot divide by $0$ , $x!=0$

The domain of $f^-1(x)$ is $D_(f^-1(x))=RR-{0}$

The range of $f(x)$ is $R_(f(x))=RR-{0}$

What is the range of the function f(x)=2/(3x-1)f(x)=2/(3x-1)?