If $f(x) = \frac{4x+3}{6x-4}, x \neq \frac{2}{3},$ show that $(f \circ f)(x) = x$ for all $x \neq \frac{2}{3}.$ What is the inverse of $f$?
(N/A) Given $f(x) = \frac{4x+3}{6x-4}, x \neq \frac{2}{3}.$
$(f \circ f)(x) = f(f(x)) = f\left(\frac{4x+3}{6x-4}\right)$
$= \frac{4\left(\frac{4x+3}{6x-4}\right) + 3}{6\left(\frac{4x+3}{6x-4}\right) - 4}$
$= \frac{\frac{16x+12 + 18x-12}{6x-4}}{\frac{24x+18 - 24x+16}{6x-4}}$
$= \frac{34x}{34} = x.$
Since $(f \circ f)(x) = x = I(x),$ the function $f$ is its own inverse.
Therefore,the inverse of $f$ is $f$ itself,i.e.,$f^{-1}(x) = f(x) = \frac{4x+3}{6x-4}.$
$(f \circ f)(x) = f(f(x)) = f\left(\frac{4x+3}{6x-4}\right)$
$= \frac{4\left(\frac{4x+3}{6x-4}\right) + 3}{6\left(\frac{4x+3}{6x-4}\right) - 4}$
$= \frac{\frac{16x+12 + 18x-12}{6x-4}}{\frac{24x+18 - 24x+16}{6x-4}}$
$= \frac{34x}{34} = x.$
Since $(f \circ f)(x) = x = I(x),$ the function $f$ is its own inverse.
Therefore,the inverse of $f$ is $f$ itself,i.e.,$f^{-1}(x) = f(x) = \frac{4x+3}{6x-4}.$
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