Inflection points introduction | AP Calculus AB | Khan Academy

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Inflection points are points where the function changes concavity, i.e. from being "concave up" to being "concave down" or vice versa. They can be found by considering where the second derivative changes signs. In similar to critical points in the first derivative, inflection points will occur when the second derivative is either zero or undefined. Created by Sal Khan.
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[@newkoreathreekingdom]

Lol thank you so much right when i needed help in inflection point 1 minute later you upload it.

[@guitarttimman]

Awesome job boss. I love the Khan Academy videos. Thank you.

[@whitneycarpenter691]

Helpful! Thank you!

[@indiawelch4592]

This helped a lot. THANKS!!!!

[@tammyl4586]

Increasing & Decreasing and Concave up & down are two different concepts...... But this video mixes them together.

[@spiguy]

The language can be confusing, but weren't they talking about depression and ascension of the slope, not the original curve?

[@Anielisstudying]

Alright, but in the third example, the functions graph represent a straight line. There is no concavity in that.
I just don’t get where the inflection point came from .

[@spiguy]

But the straight line is not an arbitrary curve, it's the 2nd derivative of the top curve. The right of the screen explains how we found the inflection point in terms of the maths. The reasoning is the same: we're looking for the critical points of f'(x) which with this method will allow us to find where the slope of f(x)'s continuity is the steepest aka inflection points.

[@jamesepace]

I think I learned that the inflection point is where the second derivative is 0. I suppose if the second derivative is always 0, then that isn't true.

[@JH-hb5cc]

James Pace not all points where the second derivative is 0 is necessarily an inflection point. All inflection points do have a second derivative of 0

[@kishansahu2518]

Manh..it's a sufficient condition

[@youtubeessentials2996]

Thanks a lot

[@deependragaur4991]

Thanks 🙏🙇

[@SPYmatholution]

Find number of inflection point for the graph f(x)= x^2, for x>=0
f(x)=-x^2 for x

[@elmarjuz]

This is not completely related to the topics you are lately showcasing, but, could you at some point turn in the future to explaining discrete math elements?
Sets and relations and stuff. You have a couple of videos about relations and functions, but these topics can be sort of convoluted.
Also, mathematical proofs. I'm having a lot of problems with those.

[@kabirkrishna8520]

Top G

[@JosePablo24]

He forgets to mention that inflection points also happen where f'(x) and f''(x) are undefined.

[@parsazahed4210]

will they be the same points? (the points where f'(x) and f"(x) are undefined)

[@aryamanm2170]

those are called critical points

[@Genghiskaran]

cool

[@vitorpareja]

does that aply for sin and cos functions ???
because the f ' (sin) isn't a cos!?!?!?

[@jjcamina5060]

Yes it applies

[@user-me6bt8gv3n]

0:28 The slope was decreasing?

[@user-me6bt8gv3n]

khan Academy are u serious

[@kabirkrishna8520]

W

[@mubasshirmubin7278]

bet u cant draw 2 concave downwards together😂😂