I'm working on an exercise from a book in the chapter on quadratic inequalities: "Find the set of possible values of the given function $\frac{x - 2}{(x + 2)(x - 3)}$". The answer in the book is "all values". I don't have much intuition on the calculations so would be grateful if someone could explain what happens at the various stages. Corrections, qualifications, etc. to my statements are very much welcomed!

Here's my workings:

Looking for the range of values of $y$, so $y = \frac{x - 2}{(x + 2)(x - 3)} \Rightarrow y = \frac{x - 2}{x^2 - x - 6}$

$\Rightarrow y(x^2 - x - 6) = x - 2$

$\Rightarrow yx^2 -(y + 1)x - 6y + 2 = 0$

As far as I can understand from my book, now that I've got a quadratic in $x$, I can use the discriminant to determine the $y$ range . With $a = y, b = -(y + 1), c = (-6y + 2)$, the discriminant ($b^2 - 4ac$) is:

$(-y-1)^2 - 4(y)(-6y + 2)$

$= 25y^2 - 6y + 1$

When $25y^2 - 6y + 1 \geqslant 0$, the roots of the quadratic in $x$ are real. At this stage, from what I understand, the values satisfying the inequality $25y^2 - 6y + 1 \geqslant 0$ are the range of $y$ values for all valid, real $x$ values plugged into the function $\frac{x - 2}{(x + 2)(x - 3)}$.

Solving the inequality using the quadratic formula with $a = 25, b = -6, c = 1$:

roots: $\frac{6\pm\sqrt{36 - (4)(25)(1)}}{50}$

$ = \frac{6\pm\sqrt{-64}}{50}$. Because the discriminant here $< 0$ then there are no real roots. I don't get how this is interpreted as "all values" (i.e. the answer in the book).

  • $\begingroup$ The calculation shows that for any real number $y$, there is a real number $x$ such that $y=\frac{x-2}{(x+2)(x-3)}$ precisely if $25y^2-6y+1 \ge 0$. But this inequality holds for all real $y$. This is because certainly $25y^2-6y+1$ is positive somewhere, and since it is never $0$ (by your calculation), it must be positive everywhere. You would learn a little more by completing the square. Your expression is equal to $(5y-3/5)^2 -(3/5)^2) +1$, so it is always $\ge 1-(3/5)^2$. $\endgroup$ – André Nicolas Apr 22 '11 at 16:04

Since there are no real roots, the graph of $25y^2−6y+1$ cannot cross the horizontal axis and has to be always negative or always positive. Since it is positive for $y=0$, it is always positive.

Since this is the expression for your original equation, this means that you can always find a solution $x$ which gives this value of $y$ (although you should check the trivial fact that this never gives a 0-denominator).

  • $\begingroup$ I get it now. Gone through the problem again and understand what each quadratic and discriminant actually means. Thanks for your succinct and clear explanation. $\endgroup$ – PeteUK Apr 23 '11 at 9:13

This answer is just to point out another way to find the range, since I believe your specific questions have been answered by user9325.

Consider $f(x)=\frac{x-2}{(x+2)(x-3)}$ on the interval $(-2,3)$, where it is everywhere defined and continuous. Since $x+2$ is always positive on this interval and $x-3$ is always negative, the sign is opposite that of $x-2$.

As $x$ approaches $3$ (from the left), $x-2$ will be positive, so $f(x)$ will be negative. The denominator approaches $0$ while the numerator approaches $1$, so the absolute value of $f$ will go to $\infty$. Thus $f(x)\to-\infty$ as $x\to 3$ from the left.

As $x$ approaches $-2$ (from the right), $x-2$ will be negative, so $f(x)$ will be positive. The denominator approaches $0$ while the numerator approaches $-4$, so the absolute value of $f$ will go to $\infty$. Thus $f(x)\to+\infty$ as $x\to-2$ from the right.

Because $f$ is continuous on $(-2,3)$ and takes on arbitrarily large positive and negative values on that interval, the range of $f$ is all real numbers by the Intermediate Value Theorem.

  • $\begingroup$ The function $f$ is a typical candidate for partial fractions. The familiar computation gives $f(x)={1\over5}({4\over x+2}+{1\over x-3})$ from which Jonas' findings can be immediately read off. $\endgroup$ – Christian Blatter Apr 22 '11 at 17:53
  • $\begingroup$ @Christian: Thanks, that is a nice way to look at it. However, how "immediate" or "familiar" it is will depend on who's doing the reading or computing :) $\endgroup$ – Jonas Meyer Apr 22 '11 at 18:02

Not sure I can directly answer your final question, since I've forgotten what I've learn about how to apply discriminants in your case. However...

It is clear from inspection of the factored form of the original "y = ..." equation, that the range of y is indeed all values of y from -infinity to +infinity, by observing the behavior of the function for x = -2 +/- dx, x = 2, and x = 3 +/- dx.

  • 1
    $\begingroup$ This alone is not enough to say that all values are assumed; only that the range must be unbounded. Consider for example $f(x)=(x^2+1)/x$, which has the line $x=0$ as a vertical asymptote, but does not assume any values between $-2$ and $2$. $\endgroup$ – Hans Lundmark Apr 22 '11 at 17:08
  • $\begingroup$ I think it does, given that I included the point for x=2; but I agree that I maybe left too much as an exercise for the reader. The value of the function approaches +infinty for x slightly more positive than -2. The value approaches -infinity for x slightly more negative than +3. The value of the function at x=2 is 0. There are no discontinuities in the range -2 < x < 3, and x goes from +infinity through 0 to -infinity in that range, so all Y are covered by the intermediate value theorem (or is it continuity theorem? I forget...). $\endgroup$ – Vintage Apr 22 '11 at 17:35
  • $\begingroup$ Oops, sorry, I didn't see that "x=2" hidden in there. The argument in your comment of course suffices, but when you're saying that the function has no discontinuities, you are implicitly looking at other points than just close to $-2$, $2$ and $3$. ;-) Anyway, +1 for the clarification, and welcome to this site! $\endgroup$ – Hans Lundmark Apr 22 '11 at 17:40

You can rewrite



$25((y - 3/25)^2 + 16/25^2)$

which is clearly > 0 for any value of y


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