# Finding Local Maximum & Minimum f(x)=x/(x^2+x+1)

• Nov 5th 2012, 12:28 AM
iFuuZe
Finding Local Maximum & Minimum f(x)=x/(x^2+x+1)
I've been having alot of trouble with this question firstly i took the derivative of it and got (1-x^2)/(x^2+x+1)^2. Now to get the critical numbers i make the top and bottom =0. 1-x^2=0, x=1 Now for (x^2+x+1)^2 do i make (x^2+x+1)=0, x=-1 & (x^2-x-1)=0, x=1.? So from that its increasing from (-infinity,1) and decreasing from (1, infinity). Thats all i got.

If any can redo the question if i made errors it will be much appriciated.
• Nov 5th 2012, 12:38 AM
MarkFL
Re: Finding Local Maximum & Minimum f(x)=x/(x^2+x+1)
You have correctly differentiated to find:

$\displaystyle f'(x)=\frac{1-x^2}{(x^2+x+1)^2}=\frac{(1+x)(1-x)}{(x^2+x+1)^2}$

The denominator has no real roots, since the discriminant is negative. So, you only need test the sign of the numerator on the intervals:

$\displaystyle (-\infty,-1),\,(-1,1),\,(1,\infty)$

From this, you can find the nature of the two extrema.