How do you find the domain and range of $f(x) = 1 / (x-5)$ ?

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Answer 1 · 2017-02-04T14:27:45

The domain is $D_f(x)=RR-{5}$
The range is $R_f(x)=RR-{0}$

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

So, the domain of $f(x)$ is $D_f(x)=RR-{5}$

Let $y=1/(x-5)$

$y(x-5)=1$

$yx-5y=1$

$yx=1+5y$

$x=(1+5y)/y$

Therefore,

$f^-1(x)=(1+5x)/x$

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

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

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

How do you find the domain and range of f(x) = 1 / (x-5)f(x) = 1 / (x-5)?