How do you find the domain and range for $f(x)=(2x+1)/(x-3)$ ?

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Answer 1 · 2017-04-29T06:39:44

The domain of $=RR-{3}$
The range of $=RR-{2}$

As we cannot divide by $0$ , $x!=3$

The domain of $f(x)$ is $D_f(x)=RR-{3}$

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

Then,

$yx-3y=2x+1$

$yx-2x=3y+1$

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

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

Therefore,

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

The domain of $x$ is the range of $y$

The range of $f(x)$ is $R_f(x)=RR-{2}$
graph{(2x+1)/(x-3) [-28.86, 28.9, -14.43, 14.43]}

How do you find the domain and range for f(x)=(2x+1)/(x-3)f(x)=(2x+1)/(x-3)?