The question asks for the critical points and inflection points of:
(x^2-1)/(x^3+1)
I've tried to take the derivitave and second derivitave but it just got too complicated. If anyone could help me that would be great.
My quick commands menu doesn't seem to be working but the question is quite simple.
Thanks!
Thanks alot for clearing up the critical points.
I'm pretty terrible at my derivitave calculation though so once again i'm having alot of trouble finding the second derivitave in order to find the inflection points.
As Quacky pointed out the first derivitave is -x(x-2)/(x^2-x+1)^2
If anyone could help me that would be great .
Well all I have to say is that there must be an easier way:
(-x^2+2x)'(x^2-x+1)^2 - (x^2+2x)(x^2-x+1)' / (x^2-x+1)^4
(-2x+2)(x^2-x+1)^2 - (x^2+2x)[2(x^2-x+1) x (2x-1)] / (x^2-x+1)^4
(-2x+2)(x^4-2x^3+3x^2-2x+1) - (x^2+2x)(4x-2)(x^2-x+1) / (x^2-x+1)^4
(-2x^5+2x^4-6x^3+4x^2-2x+2x^4-4x^3+6x^2-4x+2) - (4x^5-4x^3-4x) / (x^2-x+1)^4
And from there... I have no idea .
I'm not saying to try it haha, I know there must be another way, thanks for the help!
Sorry for the first reply where I missed '/' sign and solved (x^2-1) (x^3+1)
Here is new reply:
expression can be written as
y = ( x - 1 ) / ( x^2 - x + 1 )
then y' = [ ( x^2 - x + 1 ) 1 - ( x - 1 ) ( 2x - 1 ) ] / ( x^2 - x + 1 )^2
gives y' = ( 2x - x^2 ) / ( x^2 - x + 1 )
again y" = [ ( x^2 - x + 1 )^2 ( 2 - 2x ) - ( 2x - x^2 ) 2( x^2 - x + 1 )( 2x - 1 ) ]/ ( x^2 -x + 1 )^4
gives y" = ( 2x^3 - 6x^2 + 2 ) / ( x^2 - x + 1 )^3 and we find point of inflection