Question

Can someone explain me how y=ln(x^2+1) have two inflection points at x=1 and x=-1 instead of just one at x=0?

Answer 1

inflection points are when we change concavity, so if we take the second derivative and set it equal to 0, we may find these points.

Answer 2

can you tell me what the second derivative is?

Answer 3

@Mateaus ?

Answer 4

Yeah I know but aren't the first and second derivative to find and graph the original function?

Answer 5

There are many things we can do with the first and second derivatives. One of the things we can do with the second derivative is easily find the inflection points.

Answer 6

well easily....

Answer 7

Yeah but seeing the original function. There is just one inflection point at x=0

Answer 8

I think you are confusing vertex with inflection

Answer 9

change in concavity is where it goes from a frowny face to a smiley face

Answer 10

look at the graph I see that its around x=2 and x= -2

Answer 11

1 and -1
which is what you get when you set the second derivative equal to 0

Answer 12

http://www.wolframalpha.com/input/?i=inflection+points+y%3Dln%28x^2%2B1%29

Answer 13

look at that

Answer 14

I am aware of that

Answer 15

which part?

Answer 16

All. I am just confuse on describing the graph of the original function with the second derivative. I am aware that using the first one to find max and mins it can help me graph the original function but the second one doesn't help?

Answer 17

yes it helps, the second derivative will show you where something changes concavity, it will also tell you if the concavity is up or down

If the second derivative is positive on some interval, then the original function is concave up on that interval
If the second derivative is negative on some interval, then the original function is concave down on that interval

and so when the second derivative is equal to zero, we have a "potential" change in concavity.

Answer 18

exactly. Looking at the second derivative. It says x=1 and x=-1 are inflection points when in the original function there is none

Answer 19

I say potential because the second derivative may just bounce of the x axis, in which case it does not change concavity , and then we do NOT have a inflection point

Answer 20

the 1, and -1 ARE THE INFLECTION POINTS OF THE ORIGINAL FUNCTION

Answer 21

the inflection points of the original function occur when the 0's of the second derivative are not bounces.

Answer 22

we have two zeros on the second derivative that do not bounce (they go through the x axis) thus we have two inflection points

Answer 23

on the original function

Answer 24

to find the inflection points on the second derivative function, we would need the 4th derivative function

Answer 25

you are getting zeros and vertex points confused

Answer 26

with inflection points