View Full Version : range of x / (x^2 + x + 1)

baloo

05-21-2009, 11:06 AM

given function: y = x / (x^2 + x + 1)

would like to find the function's range (algebra only)

given solution: -1 <= y <= 1/3

how ?

Subhotosh Khan

05-21-2009, 12:14 PM

given function: y = x / (x^2 + x + 1)

would like to find the function's range (algebra only)

given solution: -1 <= y <= 1/3

how ?

In your graphing calculator - plot the function anyway - that will give you an idea regarding steps you need to take.

You'll see that the function has local maxima and minmum and those are global maximum and minima too ( show that as x? ±? you get y?0).

Find the y's at those maximum and minimum and that will give you the range.

baloo

05-21-2009, 02:36 PM

Thanks, just graphed it. Here it is - http://farm4.static.flickr.com/3369/355 ... 3fd8_o.png (http://farm4.static.flickr.com/3369/3551446985_82afcd3fd8_o.png)

I still don't see a way to do it without using calculus (x? ±? is not algebra).

Would be happy to learn an algebraic method if anybody knows one.

baloo

05-26-2009, 01:17 PM

ok guys, here is how (i think) the solution goes -

original equation is:

y = x / (x^2 + x + 1)

let's check which y-values are possible.

starting with substitution: y=m

m = x / (x^2 + x + 1)

mx^2 + mx + m = x

mx^2 + (m-1)x + m = 0

what values of m are required in order to solve the last equation (which are the y values of the original equation) ?

b^2 - 4ac >= 0 ==>

(m-1)^2 - 4m^2 >= 0

-3m^2 - 2m + 1 >= 0

m^2 + (2/3)m - 1/3 >=0

(m+1)(m-1/3) >= 0

-1 <= m <= 1/3

Subhotosh Khan

05-26-2009, 01:31 PM

That's a clever way to do that problem ---

Will have to remeber that...

Subhotosh Khan

05-26-2009, 01:31 PM

That's a clever way to do that problem ---

Will have to remeber that...

baloo

05-26-2009, 03:48 PM

oops... sorry, I had a mistake when copying from the handwriting (or: i tested you :) )

When dividing by -3 , the >= sign needs to change direction.

-3m^2 - 2m + 1 >= 0

m^2 + (2/3)m - 1/3 <=0

(m+1)(m-1/3) <= 0

-1 <= m <= 1/3

The bottom line was correct

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