• Mar 28th 2012, 12:43 PM
fran1942
Hello, I am trying to differentiate this function so I work out where it is increasing or decreasing.
What is the best procedure to do this. I tried using quotient rule but it seemed very laborious. Is there a better way ?

2x^2+x-1 / x^2+x-2

Thanks for any help.
• Mar 28th 2012, 01:10 PM
pickslides
Maybe try factoring each quadratic with the purpose of cancellation. Otherwise your stuck with the quotient rule.
• Mar 28th 2012, 01:41 PM
Also sprach Zarathustra
So our function is, $f(x)=\frac{2x^2+x-1}{x^2+x-2}$.

You need to find $f'(x)$.

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

$=\frac{(4x+1)(x^2+x-2)-(2x+1)(2x^2+x-1)}{(x^2+x-2)^2}$

Try to continue.
• Mar 28th 2012, 03:21 PM
fran1942
thank you. I have tried to continue to simplify this derived function so that I can work out it's roots.
I can see I could split it and cancel one of the (x^2+x-2) factors, but that is as far as I can get.
I really am stuck with this one. Any help proceeding would be great.
• Mar 28th 2012, 03:51 PM
skeeter
Quote:

$=\frac{(4x+1)(x^2+x-2)-(2x+1)(2x^2+x-1)}{(x^2+x-2)^2}$
as stated by AsZ, expand the numerator and combine like terms ... in other words, do the grunt work algebra and finish simplifying the numerator.
You'll end up with a quadratic in the numerator that won't factor, so you'll need to complete the square or use the quadratic formula to find where the derivative equals 0.
• Mar 28th 2012, 04:21 PM
fran1942
my grunt work gives me:
x^2-6x-1 / x^4+2x^3-3x^2-4x+4
I equated the numerator to zero to find the roots. That gave me roots at 6.16 and -0.162 however looking at a graph of 2x^2+x-1 / x^2+x-2 that cannot be correct.
I double checked but to no avail.
If anyone can spot my error that would be appreciated.
• Mar 28th 2012, 06:30 PM
skeeter
your roots for the derivative are correct, $x = 3 \pm \sqrt{10}$ ... note the relative max at $x = 3-\sqrt{10}$ and the relative min at $x = 3+\sqrt{10}$ in the graph (marked by the red diamonds)