<s>[INST] Range of $f(x) =\frac {x -1}{x^2 -2x + 3} $? Is my solution for finding the range of 
$$f(x) = \frac{x-1}{x^2 -2x + 3} $$
correct?
Since its Domain is $ \mathbb{R} $, so transforming this equation into $x$ in terms of $y$ , we get $$ yx^2 - (2y +1)x + (3y+1) = 0 $$
Now, since $ x \in \mathbb {R} $ , so its discriminant $\mathbb{D}$ must be $ \ge 0 $, so we have, $$ (2y + 1)^2 - 4y(3y + 1) \ge 0 $$
Solving it, we get
$$ y \in \left[-{1 \over {2 \sqrt2}}, {1 \over {2 \sqrt2}}\right] $$
Kindly solve my problem.
 [/INST] We can more easily find out maxima/minima to find its range:
To find where extrema occur by Quotient Rule:
$$ \frac{x-1}{x^2-2 x +3} = \frac{1}{2 x -2 } $$
Solving,
$$ x = 1 \pm \sqrt 2 $$
Extreme values are:
$$ \pm \dfrac{\sqrt 2}{4} $$
which is the range.

</s>