Consider the function $f (x) = \frac{1}{x - 2}$ $f(x)=1/(x-2)$ , then how do you simplify $f (f^{- 1} (x))$ $f(f^-1(x))$ ?

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Answer 1 · 2016-08-19T20:25:29

$f^{- 1} (x) = \frac{1}{x} + 2$ $f^{-1}(x) = 1/x+2$ and
$f (f^{- 1} (x)) = x$ $f(f^{-1}(x) ) = x$

Given $f (x) = \frac{1}{x - 2}$ $f(x) = 1/(x-2)$ then

$f (g (x)) = \frac{1}{g (x) - 2}$ $f(g(x)) = 1/(g(x)-2)$ . Now, if $f (g (x)) = x$ $f(g(x)) = x$ then

$g (x) = f^{- 1} (x)$ $g(x) = f^{-1}(x)$

but

$\frac{1}{g (x) - 2} = x$ $1/(g(x)-2)=x$ gives $g (x) = \frac{1}{x} + 2$ $g(x) = 1/x+2$

then

$f^{- 1} (x) = \frac{1}{x} + 2$ $f^{-1}(x) = 1/x+2$

Consider the function f(x)=1x−2f(x)=1/(x-2), then how do you simplify f(f−1(x))f(f^-1(x))?