point of inflections

[bar0029]

hey..

if you have a function, for example f(x) = sin(x) where x is an element of [0,2pi]
is the point of inflection at x=0, pi and 2pi or just at x=pi..
because i did a question and the answer said it was at all three points
it would make sense if it was just at x=pi, given the fact that this is the only place where the second derivative graph ACTUALLY changes sign

THANKING YOU!

[xZero]

the second derivative is f''(x) = -sin(x), f''(x) = 0 when x = 0,pi and 2pi. So its a point of inflection at all 3 points

[bar0029]

yeah, i know, but the graph doesn't actually change from negative to positive and vise versa at those points except x=pi, therefore, is it still a point of inflection?
because on the original graph it does not represent a point of inflection...

i think i may just have answered my own question

[brightsky]

yeah the concavity must change signs in order to for the point to be an inflection point. f''(x) = 0 isn't a sufficient condition but it is a necessary one.

edit: error in expression

[BubbleWrapMan]

It's x = pi only, you can't differentiate the endpoints of f(x).

[rama235]

yeah the concavity must change signs in order to for the point to be an inflection point. f''(0) isn't a sufficient condition but it is a necessary one.

isn't that for a stationary point of inflection?

[brightsky]

yeah the concavity must change signs in order to for the point to be an inflection point. f''(0) isn't a sufficient condition but it is a necessary one.

isn't that for a stationary point of inflection?

nup it's for inflection points in general. stationary points of inflection also have the derivative equal to 0.

[luffy]

yeah the concavity must change signs in order to for the point to be an inflection point. f''(0) isn't a sufficient condition but it is a necessary one.

isn't that for a stationary point of inflection?

Think of it logically. If you have a certain graph, the point of inflection is a point of maximum/minimum gradient in the local domain. Hence, the gradient function will have a local maximum/minimum at that point. Therefore, if you find the "gradient" function of the gradient function (i.e. the second derivative), it will be equal to 0 at the point of inflection.

[RobM8]

yeah the concavity must change signs in order to for the point to be an inflection point. f''(0) isn't a sufficient condition but it is a necessary one.

isn't that for a stationary point of inflection?

Think of it logically. If you have a certain graph, the point of inflection is a point of maximum/minimum gradient in the local domain. Hence, the gradient function will have a local maximum/minimum at that point. Therefore, if you find the "gradient" function of the gradient function (i.e. the second derivative), it will be equal to 0 at the point of inflection.

The second derivative must cross the x -axis either side of a point (go from negative to zero to positive or vice versa) in order for it to be a point of inflexion, not just be equal to zero at that point.

[luffy]

yeah the concavity must change signs in order to for the point to be an inflection point. f''(0) isn't a sufficient condition but it is a necessary one.

isn't that for a stationary point of inflection?

Think of it logically. If you have a certain graph, the point of inflection is a point of maximum/minimum gradient in the local domain. Hence, the gradient function will have a local maximum/minimum at that point. Therefore, if you find the "gradient" function of the gradient function (i.e. the second derivative), it will be equal to 0 at the point of inflection.

The second derivative must cross the x -axis either side of a point (go from negative to zero to positive or vice versa) in order for it to be a point of inflexion, not just be equal to zero at that point.

My post wasn't about sufficient conditions. I was pointing out why f''(x) = 0 for a point of inflection in general, as opposed to only "stationary" points of inflection.