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 $y = \frac{5}{1 - 3 x}$ $y=5/(1-3x)$ , $\forall x > 1$ $AA x>1$

Therefore,

$y (1 - 3 x) = 5$ $y(1-3x)=5$

$y - 3 x y = 5$ $y-3xy=5$

$3 x y = y - 5$ $3xy=y-5$

$x = \frac{y - 5}{3 y}$ $x=(y-5)/(3y)$

So,

$f^{- 1} (x) = \frac{x - 5}{3 x}$ $f^-1(x)=(x-5)/(3x)$

The domain of $f^{- 1} (x)$ $f^-1(x)$ 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]}