Question

Find the interval on which the curve of "picture below" is concave up

Answer 1

@Ashleyisakitty @ganeshie8 @Hero @jim_thompson5910

Answer 2

i know i have to find the second derivative im just not sure how

Answer 3

\[y=\int\limits_{0}^{x}\frac{ dt }{ 1+t+t^2 }=\int\limits_{0}^{x}\frac{ dt }{ t^2+t+\frac{ 1 }{ 4 }-\frac{ 1 }{ 4 }+1 }\]
\[=\int\limits_{0}^{x}\frac{ dt }{ \left( t+\frac{ 1 }{ 2 } \right)^2+\frac{ 3 }{ 4 } }=\int\limits_{0}^{x}\frac{ dt }{ \left( t+\frac{ 1 }{ 2 } \right)^2+\left( \frac{ \sqrt{3} }{ 2 } \right)^2 }\]
\[=\frac{ 1 }{ \frac{ \sqrt{3} }{ 2 } }\tan^{-1} \frac{ t+\frac{ 1 }{ 2 } }{ \frac{ \sqrt{3} }{ 2 } }|0\rightarrow x\]
\[=\frac{ 2 }{ \sqrt{3} }\tan^{-1} \frac{ 2t+1 }{ \sqrt{3} }|0\rightarrow\ x\]
\[=\frac{ 2 }{ \sqrt{3} }\left\{ \tan^{-1} \left( \frac{ 2x+1 }{ \sqrt{3} }\right)-\tan^{-1}\frac{ 1 }{ \sqrt{3} } \right\}\]

Answer 4

wow okay give me a second to read that...

Answer 5

@surjithayer there's no need to integrate

you just need to find \(\Large \frac{d^2y}{dx^2}\)

Answer 6

oh no
first derivative is the integrand

Answer 7

oh what @jim_thompson5910 said

Answer 8

the first derivative is not too bad because you can use the fundamental theorem of calculus

Answer 9

yea thats what im having a problem with , do i just plug in x for every where t is at and then integrate that or do i just plug in x and then that's my first derivative?

Answer 10

first derivative
replace each \(t\) in the integrand by an \(x\)
that is all
then take the derivative of the result to get the second derivative

Answer 11

interval is 0 to x or we can change it to x then diff.

Answer 12

i mean derive not integrate

Answer 13

yeah plug in x for t

Answer 14

then DIFFERENTIATE that , not integrate

Answer 15

okay okay one second let me do that

Answer 16

i also said differentiate.

Answer 17

Fundamental Theorem of Calculus (one of the 2 parts)

If \[\Large g(x) = \int_{a}^{x}f(t)dt\] then \[\Large g \ '(x) = f(x)\]

Answer 18

surjithayer you weren't incorrect. It was just a step in the opposite direction adding unnecessary steps. Then again, doing it that way helps see how the FTC works

Answer 19

sorry, i have not used that theorem.

Answer 20

thank you for reminding me.

Answer 21

okay im getting \[\frac{-(2x+1) }{ (x^2+x+1)^2}\]

Answer 22

looks good to me

Answer 23

agreed

Answer 24

i would have written the numerator as
\[-2x-1\] so that you can solve
\[-2x-1>0\] rather easily

Answer 25

okay out of curiosity whats the second part of the theorem ... it sounds very "fundamental "

Answer 26

that is that \(F(b)-F(a)\) business

Answer 27

Other part of FTC

\[\Large \int_{a}^{b}f \ '(x) dx = f(b) - f(a)\]

Answer 28

arh damn i don't know who to give the medal too you were all so helpful

Answer 29

Most books make F ' = f, which is confusing to me because there are 2 different f's to keep track of

Answer 30

thats weird every time i see F it means \[intf(x)\]

Answer 31

F ' = f is equivalent to saying "F is the indefinite integral of f"

Answer 32

notation like to be the death of us all
nothing worse than having two identical notations for different things

Answer 33

ill keep that in mind :D

thanks again every one !!

Answer 34

the reason I set up FTC 2 that way was because most calc courses start with derivatives and then teach integrals. So why not set up the integral in terms of a derivative

Answer 35

no problem

Answer 36

okay real quick so i should graph that and where its positive is the integral that the original graph is concave up right?!?

Answer 37

it is concave up where
\[-2x-1>0\]

Answer 38

Exactly. The denominator is always positive, so you can ignore that portion and focus on where the numerator is positive.

Answer 39

so it would be (-infinity , -.5) right?

Answer 40

okay ill go ahead and take that as a silent nod

Answer 41

yes when x < -1/2