Find the range of function f(x) = (x^2-3x+2)/(x^2-4x+3).

$y = f(x) = \frac{(x-1)(x-2)}{(x-3)(x-1)}=\frac{x-2}{x-3},x≠1, 3$

Now $yx - 3y = x - 2$

$∴x=\frac{3y-2}{y-1}$

Clearly y ≠ 1.

Also, when x = 1, y = 1/2

Therefore, range of function is $R - \{\frac{1}{2},1\}$