# 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 $x=f(y)$.

-
If you solve your equation for $x$, you get $x=\frac{3y-1}{y-1}$ which implies the range $y\neq1 .$ – Mhenni Benghorbal Feb 2 '13 at 7:33

$$y = \frac{x^2 - 3x + 2}{x^2 - 5x +6} =\frac{(x -1)(x-2)}{(x-2)(x-3)} = \frac {x-1}{x-3}\quad\text{x\neq 2}$$

Now, your range does not contain the value of $y = f(2)$, since $y$ is undefined there, and the value of $y = f(3)$, as $y$ is undefined there as well. That is, your function is not defined at $x = 2$ or at $x = 3$, as in those cases, the denominator of $y$ is $0$, but defined everywhere else.

There is no need to write the function as a function of $y$ to find the range. When is $y = 0$? When is $y > 0$, and when is $y \lt 0$. To see its range, it helps in cases like this, to graph the function.

Let's discuss domain first, then range:

Let's denote by $y$ the original equation (before canceling out a common factor). Here, we have $\text{Domain}\,(y) = (-\infty, 2) \cup (2, 3)\cup (3, \infty)$.

Now put $f(x) = \dfrac{x - 1}{x - 3}$, the reduced function, defined on $\text{Domain}(f) = (- \infty,3) \cup (3,\infty)$ (see graph below).

Now, let's discuss ranges:

Note, as $x$ grows increasingly large, (very very large: $\to infty$), or increasingly small ($\to \infty$), $f(x) \to 1$ in each case. That is, we have a horizontal asymptote at $y = 1$

So, $\text{Range}(f) = (- \infty,1) \cup (1,\infty)$, and you can view that below:

graph of $f(x) = \dfrac{x-1}{x-3}$, a hyperbola.

Now, if you want to determine $\text{Range}(y)$, we need to omit $f(2) = 0$ from $\text{Range}(f)$ because, as we noted above, $2 \notin \text{Domain}(y)$ and because $f$ is a one-to-one function (which means that $f$ does not attain the value $0$ anywhere else other than at $x = 2$, and it is undefined at x = 2).

Therefore, $\text{Range}(y) = (- \infty,0) \cup (0,1) \cup (1,\infty)$.

-
thanks..... but if no factors cancel out then? – user60576 Feb 1 '13 at 17:23
thanks..... but if no factors cancel out then? like if it is.....y=((x^2)-3x+2)/((x^2)-7x+12) – user60576 Feb 1 '13 at 17:30
Then you check where the denominator is $0$, where it will be undefined, if there is such a value(s). In this case $x^2 - 7x + 12 = (x - 3)(x - 4)$ and so the denominator will be $0$, and hence undefined, at $x = 3$ and at $x = 4$ – amWhy Feb 1 '13 at 17:37

Here the problem collapses, since the two quadratics have a common linear factor.

You asked in a comment, what about if there is no such simplifying technique? Consider the equation $$y=\frac{P(x)}{Q(x)},$$ where $P$ and $Q$ are of degree $\le 2$, and at least one has degree $2$. Rewrite this equation as $yQ(x)-P(x)=0$.

By gathering like powers of $x$ together, we get a quadratic equation, with coefficients that involve $y$.

For any $y$, the quadratic equation has a real solution $x$ if the discriminant ("$b^2-4ac$") is $\ge 0$. (We have to be a bit careful about the case $a=0$.)

The discriminant is a quadratic in $y$, so it is not hard to determine the values of $y$ for which it is $\ge 0$.

-

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, $\text{Range}(f) = (- \infty,0) \cup (0,1) \cup (1,\infty)$.

-