Question

consider a function f with derivative f'(x)=sin(2x^3) on the interval -1.5 12 years ago

Answer 1

Ok so I know points of inflection occur where the second derivative is equal to zero. I found the derivative of the given derivative, graphed the function and found five places on the interval where it was equal to zero. But 5 isnt the right answer. What am I doing wrong?

Answer 2

@phi

Answer 3

@sourwing

Answer 4

what 5 places did you find ?

Answer 5

x=-1.331,-.923,0,.923,1.331

Answer 6

what did you get for f'' ?

Answer 7

6x^2cos(2x^3)

Answer 8

maybe they are counting the 0 twice, and there are 6 zeros

Answer 9

Oh wait, could it be that the graph doesnt actually cross the x axix at 0? Heres the graph so you can see what I mean https://www.wolframalpha.com/input/?i=graph+6x%5E2cos%282x%5E3%29

Answer 10

the other possibility is this (from wikipedia)

If x is an inflection point for f then the second derivative, f″(x), is equal to zero if it exists, but this condition does not provide a sufficient definition of a point of inflection. One also needs the lowest-order (above the second) non-zero derivative to be of odd order (third, fifth, etc.). If the lowest-order non-zero derivative is of even order, the point is not a point of inflection, but an undulation point

However, in algebraic geometry, both inflection points and undulation points are usually called inflection points.

Answer 11

So there would be only 4 points right?

Answer 12

yes, I would go with 4

Answer 13

Alright, thanks for your help!

Answer 14

Apparently the 0 point is:

A point where the curvature vanishes but does not change sign is sometimes called a point of undulation or undulation point.