Finding range of a function Can I get the range of the following function?
$$y=\dfrac{x^2-3x+2}{x^2-5x+6}$$
I cannot isolate $x$ to get $x=f(y)$.
Thanking you in advance.

The function $ f(x) = \dfrac{x^{2} - 3x + 2}{x^{2} - 5x + 6} $ has domain $ (- \infty,2) \cup (2,3) \cup (3,\infty) $. By eliminating the common factor $ x - 2 $ in the numerator and denominator, we see that it coincides with the function $ g(x) = \dfrac{x - 1}{x - 3} $ on $ \text{Dom}(g) = (- \infty,3) \cup (3,\infty) $. Next, observe that
\begin{align}
\forall x \in (- \infty,3) \cup (3,\infty): \quad g(x)
= \frac{x - 1}{x - 3} &= \frac{(x - 3) + 2}{x - 3} \\
                      &= 1 + \frac{2}{x - 3},
\end{align}
so we get $ \text{Range}(g) = (- \infty,1) \cup (1,\infty) $, which is easily obtainable if you sketch the graph of $ g $. Then if you want to determine $ \text{Range}(f) $, simply delete $ g(2) = 0 $ from $ \text{Range}(g) $ because


*

*$ 2 \notin \text{Dom}(f) $ and

*$ g $ is a one-to-one function (this is important, because it says that $ g $ does not attain the value $ 0 $ anywhere else other than at $ x = 2 $).
Therefore, $ \text{Range}(f) = (- \infty,0) \cup (0,1) \cup (1,\infty) $.
