Thread: 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?

To prove that you have to find the domain:



I think it can't take all real values it can't take the values -3 and 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 . Rewriting will give you a quadratic equation, with y in co-efficients. Show that for every real y, the quadratic equation has real roots
. By the quadratic formula, we have . Now for x to be real always, must be non-negative for all values of y. This quadratic expression is clearly positive since the discriminant is negative.

Hello, furor celtica!

This is probably not considered a proof.

Provided that is real, prove that: . can take all real values.

We note that: . .and there are vertical asymptotes at

Part of the graph looks like this:

Code:

              :     |     :
              :     |     :*
              :     |     :
              :     |     : *
              :     |     :   *
              :     |     :      *
              :     |     :           *
- - - - - - - + - - | - - + - - - - - - -
  *          3:     :     :3
       *      :     |     :
          *   :     |     :
            * :     |     :
              :     |     :
             *:     |     :
              :     |

We see that takes all real values except 0.

And this provided by

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?

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 . Rewriting will give you a quadratic equation, with y in co-efficients. Show that for every real y, the quadratic equation has real roots
. By the quadratic formula, we have . Now for x to be real always, 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 and there is a difficulty that needs resolving if because ....

Originally Posted by mr fantastic

All values of y include and there is a difficulty that needs resolving if because ....

Yes that is true. When I divided something by 'y' in the quadratic formula, I should have been careful.

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?