Question

Determine the intervals on which the graph of F (defined on (0, infinity) is concave up.

Answer 1

\[F(x)=\int\limits_{0}^{x} \frac{ t }{ 9+t^2 }dt\]

Answer 2

Interesting they gave you an integral of a function.

Well, we know that the first derivative determines concavity, right?

Answer 3

isn't it second deriv?

Answer 4

and would I use the 2nd fundamental thm of calc for this??

Answer 5

yes, yes ...

Answer 6

Okay thx

Answer 7

The first derivative increases = concave up
The first derivative decreases = concave down

Answer 8

to get the first derivative we don't to integrate and differentiate again, do we ?

Answer 9

nope

Answer 10

so your first derivative is?

Answer 11

it would be..
\[\frac{ x }{ 9+x^2 }\]

Answer 12

right?

Answer 13

yes, it is correct!

Answer 14

Yayy!! :D

Answer 15

Now the second derivative:)

Answer 16

quotient rule?

Answer 17

Yes, you may.
You also may to re-write with negative exponent if you hate to memorize which function is differentiate first.

Answer 18

oh okay
\[\frac{ 9-x^2 }{ (9+x^2)^2 }\]

Answer 19

Yes.

Answer 20

This is negative when ?

Answer 21

so now I'd find when it's =0 for critical #s right? then do the test thingy on the number line?

Answer 22

Oh, for critical points you set the first derivative =0.

Other possible cases, in general.
Any x=a such that f(x) is defined and f'(x) is not.
Closed interval boundaries (if any),

Answer 23

Oh! Is the answer (0,3) ?? concave up?

Answer 24

It's a continuous function right, so there would be no points when it's undefined?

Answer 25

\(\displaystyle 0<\frac{ 9-x^2 }{ 9+x^2 }\)

\(\displaystyle 0<\frac{ (3-x)(3+x)}{ 9+x^2 }\)

\(\displaystyle 0<\frac{ (3-x)(3+x)}{ \rm positive }\)

So, either both parenthesis are negative or both positive.

Both negative, x<-3.
Both positive, x>3.

Answer 26

So it is concave up whenever \(x{\tiny~}\in {\tiny~~}[\infty,-3){\tiny~}\cup{\tiny~}(3,\infty)\).

Answer 27

okay thank you again :))

Answer 28

Are you sure? :o
I think it's the piece in the middle.

Answer 29

oh yeah wouldn't it be the (-3,3) ? because tht part is positive

Answer 30

Yes, you are right.

Answer 31

9 - x^2 > 0
9 > x^2
|x| < 3

Answer 32

okay so then on the interval (0,inf) the answer could only be (0,3) right?

Answer 33

stupid parenthesis thinking ... should have just done what I did now at first:)

Answer 34

lol happens to the best of us xD

Answer 35

obviously if |x|\(\ge\)3, then 9-x^2 is NOT POSITIVE.

Answer 36

Nice catch!

Answer 37

Thanks you guys!! @SolomonZelman @zepdrix

Answer 38

1 more question :'D

Answer 39

Sure

Answer 40

I'll post another one