The function f is defined by fx=5/(1-3x) for x>1 Find an expression for f^-1 and state domain and range of f^-1?

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Answer 1 · 2017-11-19T07:34:02

See the explanation below

Let $y=5/(1-3x)$ , $AA x>1$

Therefore,

$y(1-3x)=5$

$y-3xy=5$

$3xy=y-5$

$x=(y-5)/(3y)$

So,

$f^-1(x)=(x-5)/(3x)$

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

and the range is $RR-{1/3}$

If $x>1$ , then the domain is $(1,+oo)$

and

the range is

$((5/(1-3)), 0)$ , $=>$ , $(-2.5,0)$

See the curve

graph{5/(1-3x) [-10, 10, -5, 5]}