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I have the following function: \begin{equation} y=\frac{(x-1)(x+4)}{(x-2)(x-3)} \end{equation} It's easy to find it's domain, which is: $\mathbb{R} - \{3,2\}$. I know how to find the range of easier functions such as parabolas and such. Is there a systematic way to find the range of more difficult functions such as this one? Thank you in advance.

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2 Answers 2

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$y(x-2)(x-3)=(x-1)(x+4) \implies (y-1)x^2-(5y+3)x+(6y+4) = 0$. Thus for any value of $y$, there will be corresponding value(s) of $x$ iff $\Delta = (5y+3)^2-4(y-1)(6y+4) \ge 0 \iff (y+19)^2 \ge 336$, which is true for all real $y \not \in (-19-4\sqrt{21}, -19+4\sqrt{21})$.

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    $\begingroup$ It should be $(y-1)x^2+(-5y-3)x+6y+4=0$. $\endgroup$
    – mathlove
    Aug 19, 2015 at 12:22
  • $\begingroup$ @mathlove Thanks, edited it. $\endgroup$
    – Macavity
    Aug 19, 2015 at 12:28
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$$f(x)=\frac{(x-1)(x+4)}{(x-2)(x-3)}=1-\frac{6}{x-2}+\frac{14}{x-3}$$ hence the stationary points of $f(x)$ occur at $x=\frac{5\pm\sqrt{21}}{4}$.

$f$ is a continuous function on its domain and we have: $$ \lim_{x\to \pm\infty}f(x)=1,\qquad \lim_{x\to 2^{\pm}}f(x) = \mp\infty,\qquad \lim_{x\to 3^{\pm}}f(x)=\pm\infty $$ hence the range of $f$ is given by: $$ \left(-\infty,\,f\left(\frac{5+\sqrt{21}}{4}\right)\right]\cup \left[\,f\left(\frac{5-\sqrt{21}}{4}\right),+\infty\right)=\mathbb{R}\setminus\left(-19-4\sqrt{21},-19+4\sqrt{21}\right).$$

An alternative approach is the following: $f(x)=l$ is equivalent to $(x-1)(x+4)=l(x-2)(x-3)$ so a real solution exists provided that a discriminant is non-negative. Such a discriminant is $l^2+38l+25$, so we get the same conclusion as above.

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  • $\begingroup$ Thanks for answering. I did not understand the alternative approach. How did you find this discriminant $l^2+38l+25$ ? $\endgroup$
    – KeyC0de
    Aug 19, 2015 at 13:27
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    $\begingroup$ @RestlessC0bra: the discriminant of $(1-l)x^2+(3+5l)x-(6l+4)$ is just $(3+5l)^2+4(1-l)(6l+4) = l^2+38l+25$. $\endgroup$ Aug 19, 2015 at 13:48
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    $\begingroup$ Ah, so it's pretty much the same thing. I simply used $y$ instead of $l$. $\endgroup$
    – KeyC0de
    Aug 19, 2015 at 14:35

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