anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@Zarkon
anonymous
  • anonymous
https://gyazo.com/4935cc6bed8b8ff7962d6a8c115c5cc7
Yttrium
  • Yttrium
I think you can use integration using partial fractions.

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More answers

anonymous
  • anonymous
I took the second derivative.
anonymous
  • anonymous
@Yttrium you could, but the fundamental theorem of calc is better in this situation. (thanks @jim_thompson5910 ;p)
anonymous
  • anonymous
\[y''=\frac{ -12x }{ (x^2 +1)^2 }\]
anonymous
  • anonymous
@Yttrium did what?
Yttrium
  • Yttrium
Nevermind lol
anonymous
  • anonymous
Now I don't know what to do. @freckles @jim_thompson5910
jim_thompson5910
  • jim_thompson5910
I don't agree with your y '' @Ephemera you're close though
jim_thompson5910
  • jim_thompson5910
you should find that \[\Large y \ ' = \frac{6}{1+2x+x^2}\]
jim_thompson5910
  • jim_thompson5910
now use the quotient rule
anonymous
  • anonymous
That's what I did.
Yttrium
  • Yttrium
Perhaps you can also use \[y' = 6(x+1)^{-2}\] and simply use power rule.
anonymous
  • anonymous
Found my mistake
anonymous
  • anonymous
I did it with (x+1)^2 at the bottom.
anonymous
  • anonymous
Instead of that I put the original equation
anonymous
  • anonymous
And got\[\frac{ -12 }{ (x+1)^3 }\]
jim_thompson5910
  • jim_thompson5910
oh I see now, I didn't simplify fully yes i'm getting \[\Large y \ '' = \frac{-12}{(x+1)^3}\] as well
anonymous
  • anonymous
So what is needed from me now?
jim_thompson5910
  • jim_thompson5910
it asks where the function is concave up, basically it's asking for what x values makes y '' > 0 true
jim_thompson5910
  • jim_thompson5910
the numerator is fixed to being negative, so just focus on the denominator
anonymous
  • anonymous
\[x < -1\]
anonymous
  • anonymous
-1 will give you indefinite
anonymous
  • anonymous
or undefined
anonymous
  • anonymous
While positive on the bottom will give you a negative
jim_thompson5910
  • jim_thompson5910
yep, the denominator is negative when x < -1 that makes the whole right side positive
anonymous
  • anonymous
Thanks
MD152727
  • MD152727
Please close the question @Ephemera

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