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.
 A: Everything looks good, except you have a small mistake in your equation. Edit: There is no mistake anymore after OP edited his question
It should be $y x^2 - (2y + 1)x + 3y - 1 = 0$. This is solvable if and only if $$(2y + 1)^2 - 4y(3y - 1) = -8y^2 + 8y + 1 \ge 0.$$
The roots of $-8y^2 + 8y + 1 = 0$ are at
$$\frac{2 \pm \sqrt{6}}{4},$$
so we need to have $y \in \left[\frac{2 - \sqrt{6}}{4}, \frac{2 + \sqrt{6}}{4}\right]$.
A: Alternative method:
Write $f(x)$ as $\frac{x-1}{(x-1)^2+2}$. Thus, your function is a horizontal shift of $g(t)=\frac{t}{t^2+2}$, and hence these functions have the same range.
Next, $g$ is odd, so it is sufficient to study positive $t$ (for which $g$ is positive). Now, by the arithmetic-geometric mean $t^2+2\geq 2\sqrt{t^2}\sqrt{2}=2\sqrt{2}t$ with equality when $t^2=2$. Thus the maximum value is $\frac{1}{2\sqrt{2}}$. With symmetry, the minimum is $-\frac{1}{2\sqrt{2}}$. Thus, the range is (by continuity) $[-\frac{1}{2\sqrt{2}},\frac{1}{2\sqrt{2}}]$.
A: 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.

A: When the inequality is solved, I get the answer is yϵ[-1/√3,1/√3]
