Question

Intervals of concavity. If i have critical numbers x=0,-2 do i plug numbers in between them into the second derivative to find the intervals?

Answer 1

I believe this is very close to what you are asking?

http://library.thinkquest.org/3616/Calc/S2/FCPoI.html

Answer 2

Isnt that the intervals of increasing and decreasing?

Answer 3

wait, do i need all three?

Answer 4

Cant i combine them into one interval?

Answer 5

But you are given the critical numbers, so you are working backwards from the locate points of inflection right? so where they have solved for x=0,2 you are going backwards here to get your initial equations, only you are using x=0, -2

Answer 6

No, i have found both. I just dont understand how to find the intervals of concavity.

Answer 7

my inflection point(s) is x=-1 and my critical numbers are x=0,-2

Answer 8

Oh, so you are just looking for what are intervals of concavity in general?

Answer 9

I think so? i have to determine that as well as the intervals of inc and dec.

Answer 10

I know you test the numbers in between the critical numbers but i dont get how this tells me the intervals.

Answer 11

This seems to break it down a little further:

http://www.youtube.com/watch?v=Q81gafPM5y8

Does that answer it or still no?

Answer 12

I dont understand how to put the intervals together...

Answer 13

I know it changes from my inflection point and i have my number line indicating a sign change but i dont know how to put it into intervals. I tested, -3,-1,1 into the second derivative and got 4,0,-4 but i dont know where to go from there.

Answer 14

There is a full worked problem here I believe and they show at the end how to put it all together:

http://www.dummies.com/how-to/content/how-to-locate-intervals-of-concavity-and-inflectio.html

I hope that is what you are looking for?

Answer 15

it still doesnt put it into intervals....

Answer 16

I have already visited every link you posted as well.

Answer 17

ok... so have you found the 2nd derivative..?

Answer 18

so test you values x = 0 and x = -2 by substituting into the 2nd derivative

there are 3 possible outcomes

f"(x) > 0 a minimum occurs so a turning point

f"(x) < 0 a miximum occurs so a turning point

f"(x) = 0 a horizonatal point of inflection.


So if your testing results in a max or min then you should solve the 2nd derivative

let f"(x) = 0 and solve for x.

when you have the value test either side so see there is a change in sign.... hence a change in concavity.


based on your information your function may be a cubic... and should have a point of inflection at x = -1 if x =0 and x = -2 are mas and min...