1. ## Proof

If $x$ is real , prove that the expression $\frac{(x - a)(x - c)}{(x -b)}$ is capable of assuming all values provided that $a , b , c$ are either in ascending or descending orders of magnitude .

I'm lost.......how do I show that $\frac{(x - a)(x - c)}{(x -b)}$ is "capable of assuming all values".

2. You want to show that the range of this function is the set of reals. Showing it's one to one will suffice in this case since ( I am assuming)domain=co-domain=real set

3. But how do I do that ?

4. Hmm struggling myself. Obviously need to show f(x1)=f(x2) implies x1=x2 if you want to try.

5. Originally Posted by boromir
You want to show that the range of this function is the set of reals.
Right.

Showing it's one to one will suffice in this case since ( I am assuming)domain=co-domain=real set
But the function is not one-to-one, nor is the domain the set of all real numbers (x = b is not in the domain).

Try setting an arbitrary

$y=\frac{(x-a)(x-c)}{x-b},$

and see if you can find a real x that satisfies the equation, no matter what y is.

6. Here are my thoughts on this.

Let us call $f(x) = \frac{(x - a)(x - c)}{(x -b)}$

Assume that $a < b < c$. For the other case you can try it for yourself, it should be very similar. Try sketching the situation I describe below.

When $x < a$ then each of the expressions $(x-a), (x-b), (x-c)$ is negative, so f(x) is negative. As $x\rightarrow -\infty$ (meaning as x becomes a larger and larger negative number) then $f(x)\rightarrow -\infty$ (meaning the value of f(x) becomes a larger and larger negative number). Try explaining this for yourself.

At x=a, f(a)=0.

When a < x < b then f(x) is positive, which you can see by looking at whether each factor (x-a), (x-b), (x-c) is positive or negative. As x approaches b from below, the value of f(x) should approach $\infty$, in other words, f(x) should get larger and larger without bound (vertical asymptote).

This already shows that f(x) assumes all values, so you don't have to look at what goes on elsewhere (i.e. x>b), but if you're interested you can check it out yourself.
Well actually the missing piece is that the function is continuous on its domain, which simply means that there are no "breaks" in the graph.

7. consider the function $f:R\setminus\{b\} \rightarrow R$

with $f(x) = \frac{(x-a)(x-c)}{(x-b)}$. suppose $a

Consider the function on the interval $(b,\infty )$,

we take the limit $x \rightarrow b^{+}$

$\lim_{x \rightarrow b^{+}}\frac{(x-a)(x-c)}{(x-b)}= -\infty$

because $(b-a)>0$ and $b-c<0$
in the same way

$\lim_{x \rightarrow \infty}\frac{(x-a)(x-c)}{(x-b)}= \infty$

$f$ is continuos so $f$ assume all values on $R$ in the interval $(b,\infty )$

8. Rodrigo, I thought the intermediate value theorem stipulates that f has to be continuous on a closed interval for it to assume all values in it. Great work btw.

9. : )
Boromir it's a corollary of the intermediate value theorem (IVT)

If $f$ is continue in $(a, b)=I$ with $\lim_{x \rightarrow a^+} f(x) = - \infty$ , $\lim_{x \rightarrow b^-} f(x) = \infty$ so $f(I)=R$.

Take any y $\in R$ , we will show that exist $x \in (a, b)$ with $f(x)=y .$

$\lim_{x \rightarrow a^+} f(x) = - \infty$ , $\lim_{x \rightarrow b^-} f(x) = \infty$

so for any $A, B$ positive , exists $\delta_1 , \delta_2$ positive with $x \in (a, a+ \delta_1 )=I_1 \Rightarrow f(x) < -A$,

$x \in ( b- \delta_1, b )=I_2 \Rightarrow f(x) > B$, we can take $A$ and $B$ with $y \in (-A , B)$ ,

exist $\underbrace{x_1}_{\in I_1}<\underbrace{x_2}_{\in I_2}$ with

$f(x_1)<-A$ , $f(x_2)>B$ , $f(x_1) then
exist in $[x_1,x_2]$ a number $x_3$ with $f(x_3)= y$ by the IVT .

If the interval is $(a, \infty )$ with $\lim_{x \rightarrow \infty} f(x) = \infty$ so for all $B_2>0$ exists $B_1>0$

with $x>B_1$ and $f(x)>B_2$. Taking $B_2 >y$ exist $x_2> B_1$, with $f(x_2)>B_2 >y$ we use the same argument that we used before in the interval $[x_1, x_2 ] .$

10. Must say my friend (mrfour44) gave me the idea with the discriminant. I see it's probably along the lines of what Ackbeet proposed. Just consider this method because it's easier (i.e. more elementary).
So let's say our function is equal to an arbitrary real number, we have to prove that is possible. But by setting it equal to something we make it an equation and just have to show that it has a solution for real x.
So, let $a,b,c,R \in R$, a,b,c are in an ascending or descending order (i.e. $a>b>c$ or $a) and:

$\frac{(x-a)(x-c)}{x-b}=R$

Multiply by denominator and then just treat it like a quadratic:

$x^2-(a+c)x+ac=Rx-Rb$
$x^2-(a+c+R)x+ac+Rb=0$

Now, I guess you probably know what the discriminant is. So we have said that this equation holds for some real x. Now we must prove that, and this amounts to proving that the discriminant is greater than or equal to zero. So we have:

$(a+c+R)^2-4(ac+Rb)\geqslant 0$
$(a+c)^2+2R(a+c)+R^2-4ac-4Rb\geqslant 0$
$a^2+2ac-4ac+c^2+2R(a+c)+R^2\geqslant 4Rb$
$(a-c)^2+2R(a+c)+R^2\geqslant 4Rb$

Now, use the AG-GM inequality ( $a+b\geqslant 2\sqrt{ab}$) on (a-c)^2 and R^2:

$(a-c)^2+R^2\geqslant2\sqrt{(a-c)^2R^2}=2|a-c||R|$

And combine with the initial inequality:

$(a-c)^2+R^2+2R(a+c)\geqslant 2|a-c||R|+2R(a+c)\geqslant4Rb$

The rest of it is just checking whether this weaker ineqaulity is true for all the possibilities and those are (obviously it's true for R=0):

1. $a>b>c$ and $R>0$
2. $a>b>c$ and $R<0$
3. $a and $R>0$
4. $a and $R<0$

Now I won't go into that. This proof is long anyway. But this part is (considering this whole problem) rather trivial.