Question

Question on concavity...the question is atached as a picture

Answer 1

Find the second derivative of the function

Answer 2

ok hold one lemem try

Answer 3

its 3x^2 -4

Answer 4

now what

Answer 5

set it equal to zero to find the inflection point

Answer 6

its 2/3 radical 3

Answer 7

the inflectiobn point

Answer 8

is it c or d

Answer 9

what is f''(0)?
what is f''(-2)?
what is f''(2)?

each value of x is in a particular interval.

if f(x)<0 in that interval the graph is concave down.
if f(x)>0 in that interval the graph is concave up.

plug in the values and determine which intervals have f(x)<0, and which have f(x)>0

Answer 10

btw the inflection points are {-2(sqrt3)/3,2(sqrt3)/3} because square roots give two answers.

Answer 11

so is the asnwer c or d

Answer 12

:/

just plug in f''(0) and you should be able to tell whether it is c or d

Answer 13

i got d

Answer 14

what is f''(0)
tell me what number you get

Answer 15

i got f''(o) is 0

Answer 16

you already determined that
f''(x)=3x^2-4
so
f''(0)=????

Answer 17

-4

Answer 18

yes

so that means f''(0)<0 so f(x) is concave down at x=0

x=0 is in the interval (-2/3sqrt3,2/3sqrt3) so that means f(x) is concave down in that interval.

now which answer says f(x) is concave down in the interval
(-2/3sqrt3,2/3sqrt3)?

Answer 19

c

Answer 20

YES! very good. isn't it more satisfying to see WHY the answer is correct, rather than just be given it?

Answer 21

YESS!! it is but sometimes ppl dont explain it well to me so tahts y i just ask 4 teh answer...u on the other hand explained it very well

Answer 22

=) thank u

Answer 23

I try, thanks. and you're welcome :)

Answer 24

calc bc is confusing =P....teacher goesss waaaay 2 fast

Answer 25

im not dumb in matth....just need sum1 to explain it well tahts all

Answer 26

It all depends on the teacher. calc can be pretty easy if you take it step by step. You understood my explanation, so you are clearly capable of doing the work, you most likely just haven't had it explained properly. Some teachers suck :/

Answer 27

can i ask y a q....in teh beg..how did u determine the intervals to use like u needed to find f''(0) , f''(-2) and f''(2)

Answer 28

The inflection points tell the intervals. They occur when f''(x)=0

f''(x)=3x^2-4=0--->x={-2/3sqrt3,2/3sqrt3}

so those points are where the graph (potentially) changes from concave up to concave down. This divides our function into three intervals that we need to check:\[(-\infty,-2\sqrt3/3) \]\[(-2\sqrt3/3,2\sqrt3/3) \]\[(2\sqrt3/3, \infty)\]
so pick a number from each interval and plug it into f''(x)

I chose x=-2 because it is in the first interval. I could have chosen x=-5 because it is also in that interval, or any x<2/3sqrt3

similarly I chose 0 because it is in the second interval. again I had other choices, 0 just seemed easiest

and of course the same goes for why I chose 2. it's a (practically) random number that is in the third interval. It could have been any x>2/3sqrt3

I hope that helps your understanding.

Answer 29

ohhh i see now... just pick numbers within the interval to see teh behavior of the line

Answer 30

thanks for your explanation

Answer 31

Yes, that is a good way of putting it. I believe you understand. Congrads :)

Answer 32

yes it feels gud to understand math =)