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Consider the function $f(x) = (x^2-3x+2)/(x^2+x-6)$

Now I want to find the range of this function I put RHS equal to y and form a quadratuc in x and find the discriminant and get a perfect square hence for any real y x is real but the actual ramge of function does not include 1/5 and 1 so what is wrong with my method?

I know the domain of function does not include 2 and -3 but how do I know by not including these two what value of y is not possible ?

Is there a simple method that can directly help us to find the real numbers that function cannot obtain without using calculus nor drawing graph of the algebraic function as both these processes are generally very lengthy ?

Matt
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  • Factor and cancel out the $(x-2)$ on both top and bottom. You'll see why you don't have $1/5$. The 1 comes from the fact that you can divide the bottom from the top and have $1 - \frac{4x+8}{x^2 + x-6}$ – Yunus Syed May 09 '17 at 05:30
  • I dont see it and you are not dividing you are adding and subtracting 1 – Matt May 09 '17 at 05:32
  • What is wrong with my method why do I get a perfect square ? – Matt May 09 '17 at 05:39
  • Typo: $-8$ on top of fraction, not $+8$. Have you learned polynomial long division? You get the perfect square because the quadratics are nice and have two roots. – Yunus Syed May 09 '17 at 05:48
  • I dont think you got me see this https://math.stackexchange.com/q/1414298/315064 – Matt May 09 '17 at 05:58
  • Notice that the denominator in the linked example does not real-valued zeros, which is not the case here. – N. F. Taussig May 09 '17 at 09:27

1 Answers1

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What you did was that you transformed the original equation you made

$$\frac{x^2-3x+2}{x^2+x-6}=y \qquad \qquad (1)$$

by multiplying both sides with ${x^2+x-6}$, resulting in

$$x^2-3x+2=y(x^2+x-6) \qquad \qquad (2)$$.

Now that transformation logically works only in one direction: $(1) \implies (2)$. It does not work the other way around, because if $x^2+x-6=0$ then (2) can be true but (1) isn't (it isn't even defined in that case, which you noted). So you have to remember that any solutions of (2) are only valid for (1) if $x \neq 2,-3$.

Now you probably rearranged (2) to make it a quadratic in $x$:

$$(1-y)x^2-(3+y)x+(2+6y)=0 \qquad \qquad (3) $$.

(3) is only a quadratic if $(1-y) \neq 0$, which you need to consider in another case. Now you solved the quadratic. I'm omitting the details, but the end result is

$$x_{1,2}=\frac{3+y \pm(5y-1)}{2(1-y)} \qquad \qquad (4) $$.

OK, you did everything correctly except to take care of the 'little exceptions'. First, let's consider (3). You did not correctly consider $y=1$, in which case it becomes

$$-4x+8=0$$

with solution $x=2$.

So (3) actually has a solution in $x$ for each $y$, it is given by $x=2$ for $y=1$ and by (4) otherwise.

(3) is equivalent to (2), so we now need to consider the equivalence of (2) and (1). As already noted, they are not equivalent if $x=2,-3$, so we have to find which values of $y$ have only solutions in $\{2,-3\}$.

The first that springs to mind is our special case $y=1$, which has only solution $x=2$, which is thus not a solution of (1). Otherwise, we need to use (4), rembering that now $y \neq 1$:

$$\frac{3+y \pm(5y-1)}{2(1-y)}=2 \iff$$ $$3+y \pm(5y-1)=4(1-y) \iff$$ $$\pm(5y-1)=1-5y \qquad \qquad (5)$$

If we take the $+$-sign in (5) we get $y=\frac15$ as solution, with the $-$-sign all real $y$ are solutions. That means that (4) produces one 'invalid' solution $x=2$ for all $y$, and for $y=\frac15$ it is also the other solution, so $y=\frac15$ has no solution in (1).

The case $x=-3$ is solved the same way:

$$\frac{3+y \pm(5y-1)}{2(1-y)}=-3 \iff$$ $$3+y \pm(5y-1)=6(y-1) \iff$$ $$\pm(5y-1)=5y-9 \qquad \qquad (6)$$

If we take the $+$-sign in (6), there are no $y$ that satisfy it. If we take the $-$-sign, we get $y=1$, but we already know that that case cannot be discussed here.

So the final result is: (1) has a solution for every $y$ except for $y=1$ and $y=\frac15$. $y=1$ is a special case because (3) is not a quadratic in this case and $y=\frac15$ is the only case where (4) produces 2 solutions for $x$ that both are in $\{2,-3\}$, which is forbidden in (1).

Ingix
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  • After equation 5 you have written taking negative sign in we get that all y produces x as 2 so doesn't that mean that there is no y in range as all require x to be 2. In other words how did you conclude from there that y = 1/5 is not a solution whereas rest real y are. – Matt May 10 '17 at 17:11
  • I really appreciate your detailed answer definitely a plus one! It really clarified almost all my doubts ! – Matt May 10 '17 at 17:13
  • In order for any given $y$ to be in the range of $f$, there needs to be one $x$ with $f(x)=y$. The quadratic equation (3) gives you two solutions for any $y$. The analysis leading to (5) shows that for any $y \neq 1$ one of the solutions $x$ produced by (4) (the one with the "$-$"-sign) is the 'invalid' $x=2$. But the other solution (the one with the "$+$"-sign) is still valid except for $y=\frac15$, and one solution that actually carries through to (1) is enough. – Ingix May 11 '17 at 06:57