Given $f(x)=1/(x-2)$ , how do you find f(f^-1(x))?

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Answer 1 · 2016-03-09T14:59:31

First we must find the inverse.

The inverse can be found algebraically by switching y with x and vice versa.

$y = 1/(x - 2)$

$x = 1/(y - 2)$

$x(y - 2) = 1$

$y - 2 = 1/x$

$y = 1/x + 2$

We must now plug this into function ƒ, as the notation $ƒ(ƒ^-1(x))$ denotes.\

$ƒ(x) = 1/(1/x + 2)$

So, $ƒ(ƒ^-1(x)) = 1/(1/x + 2)$

Given f(x)=1/(x-2)f(x)=1/(x-2), how do you find f(f^-1(x))?