maths anyone?

i know this is a long shot, but you guys are geeks...
AS level maths, quadratics:

Find the set values taken by
_____1_____
9X2 +12X+7

For real values of X
According to the back of the text book, the answer is 0<f(x)<1/3, my problem is how you get there. any ideas?

^ the little 2 is meant to be a squared sign

 

To resolve this equation, you need to be familiar with imaginary units.

http://209.85.48.8/228/109/upload/p4466068.gif

Since Δ < 0, then there are no real roots. Rather, there are two distinct (non-real) complex roots, which are complex conjugates of each other: http://upload.wikimedia.org/math/3/2/1/321b9d2508b3c9a9f0cdd50eb598390e.png and http://upload.wikimedia.org/math/2/5/a/25aa1df2d3eacddae23aa5d90053e934.png where i is the imaginary unit.More:
http://en.wikipedia.org/wiki/Quadratic_equation
http://en.wikipedia.org/wiki/Imaginary_unit

 

Broni, that isn't quite what the question is after.

Basically it is asking what is the range of f(x) where f(x)=1/(9x^2+12x+7).

A possible solution goes along the following lines:

To answer this you prove that f(x) > 0 for all x and prove that the maximum of f(x) is 1/3.

For the maximum of f(x) you need to know the turning point of the equation, the f(x) value corresponding to where 1/9x^2+12x+7 is a maximum.
To solve this part you differentiate the equation to find the extrema.

Let g(x) = 9x^2+12x+7 then g'(x) = 18x+12 which is at min or max when g'(x)=0, i.e., 18x+12 =0 thus x=-12/18=-2/3 and this is a minimum. Therefore this point becomes a maximum in f(x), since f(x) = 1/g(x). Hence the maximum of f(x) is f(-2/3) = 1/3 (just plug the number in the f(x) equation and calculate.

For the first part you note that g(x) is at a minimum for x=-2/3 and has the value of 3 and all other values of g(x) are greater than this, i.e., always positive up to infinity g(x) >= 3. Translating this in terms of f(x), it means that f(x) is always positive but less then 1/3, i.e., greater than zero and equals zero at infinity.

Now, as differentiation is being used to solve the problem, you may need to prove that f(x) is continuous over the full range of x values. You observe that g(x) is continuous and never has the value of zero therefore f(x) is continuous as the value 1/0 (a discontinuity in 1/) never occurs.

 

And I thought that women were hard to understand!

 

Hahahaha....

 

The only thing I understand about this thread is how to portray A squared (A²). :-D

Using the numeric keypad (not the numbers above the alphabetical keys, hold down the ALT key and type in 0178 and you get ²

Bazza

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@broni: Thank you for having a go! that was where i got to, but we haven't covered imaginary numbers in class they're pretty awesome though - did some work with them with a different teacher who had the attitude of 'screw the curriculum you should learn interesting maths' (like proofs that show 2=1 and imaginary numbers):-D

@maxwell: yay :celebrate thank you

@Novice... I got what these guys^ were on about but i guess you're right it is pretty intense, hence my need for assistance
@bigbazza: cheers :-D i knew there would be a way. but on a laptop without numberpad? i'll leave you to puzzle over it?

 

There is a way to do it, but you'll have to read the manual. :-D :-D

There is probably a "function key" that turns part of your normal keyboard into a "numeric keypad" when you press/hold it, or turn it on.

In an old Compaq laptop I was given, it uses the 7,8,9 and 0 (numeric) keys, plus U,I and O (Alpha) and J.K. and L keys as 1-9 plus zero

A better way is to use a desktop keyboard (and a separate mouse) if you don't move your laptop about much. A lot easier than laptop keyboards and inbuilt "mouse". IMO.

Bazza

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At least we had some good laugh here

 

a=b
a²=ab
a²-b²=ab-b²
(a-b)(a+b)=b(a-b)
a+b=b

sub in a=b

b+b=b
2=1 :-D

go figure :-D

 

Sorry, my non-calculator math is limited to proving 2+2=/=5

 

You can't do that :eek as a-b is zero and you are dividing by zero to get a+b=b