# New guy needs help

1. Aug 20, 2005

### Tau

Hello everyone.
I am Jeremy and I hail from Trinidad. I am in my mid thirtys and currently a network administrator for a shipping firm (import/exports).

After passing through one hell of a divorce I decided to pursue my degree in Physics. I was a math geek back in seconday school and had a knack for Physics so getting this degree should prove to rekindle an old flame.

The first step in this long journey is to get 3 A levels (University of Cambridge Advance Levels). The three A levels I have to get are as follows: Mathematics, Physics and Futher Mathematics. I am now self studying the first two.

I need your help to enable me achieve my goal. I can do most of the work myself but there are some things that escape my understanding. Thus, whatever input you guys can give me is well appreciated.

(This is an A level question; I did not know weather I should have placed this post in the College section or if A levels and grade 12 are at the same skill level- please forgive me if I posted this in the wrong place)

Here is a problem that I couldn't figure out:

Provided that x is real, prove that the function [2(3x+1)]/[3(A-9)]
can take all real values. (A = x squared)

Again, thank you for your help and support.

Jeremy

Oh yeah I have 10 gmail invites to give away. Just ask and I will email it to you.

Last edited: Aug 20, 2005
2. Aug 20, 2005

### Tide

Can you factor

$$x^2-9$$

?

3. Aug 20, 2005

### marlon

Just make a sign chart of each term of this function and multiply the signs. beware that x =3 and -3 will yield infinities

marlon

4. Aug 20, 2005

### Tau

I can factorize the difference of 2 squares.

What is a sign chart?

5. Aug 20, 2005

### marlon

I am sure you know this...

suppose you have y = 2x - 2

for x = 1, y = 0

Since the slope 2 is positive this function rises as x rises so this implies

1) x < 1 => y < 0

2)x > 1 => y > 0

That is a sign chart...Just do this for each term after having factorized the x² term

marlon

6. Aug 20, 2005

### LeonhardEuler

You can show that this function takes on any value by showing that there exists some value of x that results in each value of of the function(call it y). This can be done by finding the inverse function:
$$y = \frac{2(3x+1)}{3(x^2 - 9)}$$
$$3yx^2-27y = 6x+3$$
$$3yx^2 - 6x -27y -3 = 0$$
$$x = \frac{6 +/- \sqrt{36 + 322y^2 +36y}}{6y}$$
This inverse function(s) is defined for all real values of y, except 0, so there must be at least one value of x that yields every real value of y, except, possibly, 0. However, x=-1/3 gives 0 for the function, so that completes all real values.

By the way, if you want to know how to write equations, look at this https://www.physicsforums.com/showthread.php?t=8997

7. Aug 20, 2005

### Tau

Ahhhhhhhhhhhhhhhhh
the satisfaction that one gets when he finally knows the solution to a complex problem.

Thanks Leon

Last edited: Aug 20, 2005
8. Aug 21, 2005

### HallsofIvy

Staff Emeritus
The satisfaction of having solved it yourself is even greater!