# Concavity

• Oct 8th 2008, 01:45 PM
sfgiants13
Concavity
I tried taking the derivative and it is an absolute mess, so I thought there had to be an easier way. It says find the interval on which the integral of y=1/(1+t+t^2) from 0 to x is concave upward. Is there an easier way than taking the derivative of that?
• Oct 8th 2008, 01:52 PM
p00ndawg
Quote:

Originally Posted by sfgiants13
I tried taking the derivative and it is an absolute mess, so I thought there had to be an easier way. It says find the interval on which the integral of y=1/(1+t+t^2) from 0 to x is concave upward. Is there an easier way than taking the derivative of that?

raise the whole bottom to the top so that it becomes raised to the -1.

then do chainrule, this should work.
• Oct 8th 2008, 02:07 PM
sfgiants13
I did that and I may have got it...I got all values of x, and got the derivative of that to be 1/2sqrt[1+t+t^2]. Can anyone confirm this?
• Oct 8th 2008, 02:15 PM
11rdc11
\$\displaystyle -(1+t+t^2)^{-2}(1+2t)\$
• Oct 8th 2008, 02:24 PM
p00ndawg
Quote:

Originally Posted by sfgiants13
I did that and I may have got it...I got all values of x, and got the derivative of that to be 1/2sqrt[1+t+t^2]. Can anyone confirm this?

you forgot a key part, you need to multiply by derivative of the inside. So no not really.