1. ## last of inequalities

provided that x is real, prove that the function (2(3x-1))/(3(x^2-9) can take all real values. i believe this can be done by proving that there are no real roots to the quadratic equation obtained by introducing y=f(x). but why?

2. To prove that you have to find the domain:

$\displaystyle 3(x^2-9) \ne 0 <=> x\ne -3 \wedge x\ne 3$

I think it can't take all real values it can't take the values -3 and 3

3. Originally Posted by furor celtica
provided that x is real, prove that the function (2(3x-1))/(3(x^2-9) can take all real values. i believe this can be done by proving that there are no real roots to the quadratic equation obtained by introducing y=f(x). but why?
Hmm ... not really...Its some other quadratic equation....

Let $\displaystyle y = \frac{6x-2}{3x^2-27}$. Rewriting will give you a quadratic equation, with y in co-efficients. Show that for every real y, the quadratic equation has real roots
$\displaystyle y = \frac{6x-2}{3x^2-27} \implies 3yx^2 - 6x + 2 - 27y = 0$. By the quadratic formula, we have $\displaystyle x = \frac{6 \pm \sqrt{54y^2 - 4y + 36}}{6y}$. Now for x to be real always, $\displaystyle 54y^2 - 4y + 36$ must be non-negative for all values of y. This quadratic expression is clearly positive since the discriminant is negative.

4. Hello, furor celtica!

This is probably not considered a proof.

Provided that $\displaystyle x$ is real, prove that: .$\displaystyle f(x) \:=\: \frac{2(3x-1)}{3(x^2-9)}$ can take all real values.

We note that: .$\displaystyle \lim_{x\to\infty}f(x) \:=\:0$ .and there are vertical asymptotes at $\displaystyle x = \pm3$

Part of the graph looks like this:
Code:
              :     |     :
:     |     :*
:     |     :
:     |     : *
:     |     :   *
:     |     :      *
:     |     :           *
- - - - - - - + - - | - - + - - - - - - -
*          3:     :     :3
*      :     |     :
*   :     |     :
* :     |     :
:     |     :
*:     |     :
:     |

We see that $\displaystyle f(x)$ takes all real values except 0.

And this provided by $\displaystyle f\left(\tfrac{1}{3}\right) \:=\:0.$

5. sorry i made a little mistake in the problem, so im rewriting it and giving my proof, i'd like to know if it makes sense.
f(x)=y=2(3x+1) / 3(x^2 - 9)
=> 3x^2y - 6x - 27y -2
=> 36 - 12y(-27y -2) >= 0
=> 27y^2 + 2y + 3 >= 0
=> 27y^2 + 2y + 3 has no real roots
f(x) can take on all real values

is this proof conclusive?

6. Originally Posted by furor celtica and edited by Mr F
sorry i made a little mistake in the problem, so im rewriting it and giving my proof, i'd like to know if it makes sense.
f(x)=y=2(3x+1) / 3(x^2 - 9)

=> 3x^2y - 6x - 27y -2 = 0.

Discriminant = 36 - 12y(-27y -2)

= 12(27y^2 + 2y + 3).

But 27y^2 + 2y + 3 > 0 for all values of y (why?)

Therefore f(x) can take on all real values (why?) except possibly for y = 0 (why?).

[snip]
See above for my non-definitive edit of your solution. You still need to:

1. Explain where each step comes from.

2. Clearly demonstrate that discriminant > 0.

3. Clearly explain why it follows from discriminant > 0 that y can take on on all real values except possibly for y = 0.

4. Explain why y = 0 is possible.

Originally Posted by Isomorphism
Hmm ... not really...Its some other quadratic equation....

Let $\displaystyle y = \frac{6x-2}{3x^2-27}$. Rewriting will give you a quadratic equation, with y in co-efficients. Show that for every real y, the quadratic equation has real roots
$\displaystyle y = \frac{6x-2}{3x^2-27} \implies 3yx^2 - 6x + 2 - 27y = 0$. By the quadratic formula, we have $\displaystyle x = \frac{6 \pm \sqrt{54y^2 - 4y + 36}}{6y}$. Now for x to be real always, $\displaystyle 54y^2 - 4y + 36$ must be non-negative for all values of y. This quadratic expression is clearly positive since the discriminant is negative.
All values of y include $\displaystyle y = 0$ and there is a difficulty that needs resolving if $\displaystyle y = 0$ because $\displaystyle y = 0 \Rightarrow x = \frac{6 \pm 6}{0}$ ....

7. Originally Posted by mr fantastic
All values of y include $\displaystyle y = 0$ and there is a difficulty that needs resolving if $\displaystyle y = 0$ because $\displaystyle y = 0 \Rightarrow x = \frac{6 \pm 6}{0}$ ....
Yes that is true. When I divided something by 'y' in the quadratic formula, I should have been careful.

8. but is the conclusion correct for all that? and i am not too sure myself of the connection between the 'no real roots' and the conclusion that 'all values can be taken', its just that way in my textbook and i dont have a teacher. can someone suggest a link?