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

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Answer 1 · 2017-02-14T07:21:39

The domain is $=RR-{-1}$
The range is $=RR-{0}$

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

The domain of $f(x)$ is $D_f(x)=RR-{-1}$

To find the range, we need to calculate $f^-1(x)$

Let $y=1/(x+1)$

$x+1=1/y$

$x=1/y-1$

$x=(1-y)/y$

Therefore,

$f^-1(x)=(1-x)/x$

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

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

The range is $R_y=RR-{0}$

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