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(A) To find the inverse of the function $f(x) = \frac{2x + 1}{1 - 3x}$,we set $y = f(x)$.$y = \frac{2x + 1}{1 - 3x}$Multiply both sides by $(1 - 3x)$:

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$\frac{x - 1}{3x + 2}$

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(A) To find the inverse of the function $f(x) = \frac{2x + 1}{1 - 3x}$,we set $y = f(x)$.$y = \frac{2x + 1}{1 - 3x}$Multiply both sides by $(1 - 3x)$:$y(1 - 3x) = 2x + 1$$y - 3xy = 2x + 1$Rearrange the terms to isolate $x$:$y - 1 = 2x + 3xy$$y - 1 = x(2 + 3y)$$x = \frac{y - 1}{3y + 2}$Since $f^{-1}(y) = x$,we have $f^{-1}(y) = \frac{y - 1}{3y + 2}$.Replacing $y$ with $x$,we get $f^{-1}(x) = \frac{x - 1}{3x + 2}$.

Let the function $f$ be defined by $f(x) = \frac{2x + 1}{1 - 3x}$,then $f^{-1}(x)$ is

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