Question

Find the interval on which the curve of y equals the integral from 0 to x of 2 divided by the quantity 1 plus2 times t plus t squared, dt is concave up.

Answer 1

@Zarkon

Answer 2

https://gyazo.com/4935cc6bed8b8ff7962d6a8c115c5cc7

Answer 3

I think you can use integration using partial fractions.

Answer 4

I took the second derivative.

Answer 5

@Yttrium you could, but the fundamental theorem of calc is better in this situation. (thanks @jim_thompson5910 ;p)

Answer 6

\[y''=\frac{ -12x }{ (x^2 +1)^2 }\]

Answer 7

@Yttrium did what?

Answer 8

Nevermind lol

Answer 9

Now I don't know what to do. @freckles @jim_thompson5910

Answer 10

I don't agree with your y '' @Ephemera

you're close though

Answer 11

you should find that

\[\Large y \ ' = \frac{6}{1+2x+x^2}\]

Answer 12

now use the quotient rule

Answer 13

That's what I did.

Answer 14

Perhaps you can also use \[y' = 6(x+1)^{-2}\] and simply use power rule.

Answer 15

Found my mistake

Answer 16

I did it with (x+1)^2 at the bottom.

Answer 17

Instead of that I put the original equation

Answer 18

And got\[\frac{ -12 }{ (x+1)^3 }\]

Answer 19

oh I see now, I didn't simplify fully

yes i'm getting \[\Large y \ '' = \frac{-12}{(x+1)^3}\] as well

Answer 20

So what is needed from me now?

Answer 21

it asks where the function is concave up, basically it's asking for what x values makes y '' > 0 true

Answer 22

the numerator is fixed to being negative, so just focus on the denominator

Answer 23

\[x < -1\]

Answer 24

-1 will give you indefinite

Answer 25

or undefined

Answer 26

While positive on the bottom will give you a negative

Answer 27

yep, the denominator is negative when x < -1

that makes the whole right side positive

Answer 28

Thanks

Answer 29

Please close the question @Ephemera