Answer
a) Find the inverse function
$f^{-1}(x) = \frac{3x+1}{x-2}$
b) Verify that it is the correct inverse function
$f(f^{-1}(x)) = x$
$f(\frac{3x+1}{x-2}) = \frac{2(\frac{3x+1}{x-2}) + 1}{\frac{3x+1}{x-2}-3}$
$= \frac{\frac{6x+2}{x-2} + 1}{\frac{3x+1-3(x-2)}{x-2}}$
$= \frac{\frac{6x+2+x-2}{x-2}}{\frac{3x+1-3x+6}{x-2}}$
$=\frac{\frac{7x}{x-2}}{\frac{7}{x-2}}$
$= \frac{7x}{x-2} \times \frac{x-2}{7}$
$=\frac{7x(x-2)}{7(x-2)}$
$= \frac{7x}{7}$
$= x$
$f^{-1}(f(x)) = x$
$f^{-1}(\frac{2x+1}{x-3}) = \frac{3(\frac{2x+1}{x-3})+1}{\frac{2x+1}{x-3}-2}$
$= \frac{(\frac{6x+3}{x-3})+1}{\frac{2x+1}{x-3}-2}$
$= \frac{(\frac{6x+3+x-3}{x-3})}{\frac{2x+1-2(x-3)}{x-3}}$
$= \frac{(\frac{7x}{x-3})}{\frac{2x+1-2x+6}{x-3}}$
$= \frac{(\frac{7x}{x-3})}{\frac{7}{x-3}}$
$= \frac{7x}{x-3} \times \frac{x-3}{7}$
$= \frac{7x(x-3)}{7(x-3)}$
$= \frac{7x}{7}$
$= x$
Work Step by Step
a) Find the inverse of the function
Let $f(x) = y$
$f(x) = \frac{2x+1}{x-3}$
$y = \frac{2x+1}{x-3}$
$x = \frac{2y+1}{y-3}$
$x(y-3) = 2y + 1$
$xy - 3x = 2y + 1$
$xy - 2y = 3x + 1$
$y(x - 2) = 3x + 1$
$y = \frac{3x+1}{x-2}$
Therefore, $f^{-1}(x) = \frac{3x+1}{x-2}$
b) Verify that the equation is correct
$f(f^{-1}(x)) = x$
$f(\frac{3x+1}{x-2}) = \frac{2(\frac{3x+1}{x-2}) + 1}{\frac{3x+1}{x-2}-3}$
$= \frac{\frac{6x+2}{x-2} + 1}{\frac{3x+1-3(x-2)}{x-2}}$
$= \frac{\frac{6x+2+x-2}{x-2}}{\frac{3x+1-3x+6}{x-2}}$
$=\frac{\frac{7x}{x-2}}{\frac{7}{x-2}}$
$= \frac{7x}{x-2} \times \frac{x-2}{7}$
$=\frac{7x(x-2)}{7(x-2)}$
$= \frac{7x}{7}$
$= x$
$f^{-1}(f(x)) = x$
$f^{-1}(\frac{2x+1}{x-3}) = \frac{3(\frac{2x+1}{x-3})+1}{\frac{2x+1}{x-3}-2}$
$= \frac{(\frac{6x+3}{x-3})+1}{\frac{2x+1}{x-3}-2}$
$= \frac{(\frac{6x+3+x-3}{x-3})}{\frac{2x+1-2(x-3)}{x-3}}$
$= \frac{(\frac{7x}{x-3})}{\frac{2x+1-2x+6}{x-3}}$
$= \frac{(\frac{7x}{x-3})}{\frac{7}{x-3}}$
$= \frac{7x}{x-3} \times \frac{x-3}{7}$
$= \frac{7x(x-3)}{7(x-3)}$
$= \frac{7x}{7}$
$= x$
