If f(x) is a twice differentiable function: the number of points

Does this hold up?

Concavity changes at relative extrema.

>>8522371
Concavity changes at points of inflection.

>>8522336
yes this is true

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>>8523559
Prove it

f(x) = x^2
lmao fuck off OP

>>8523583
fuck me im retarded

>>8523583
Theres one minimum in that function, Therfore there are 1-1=0 infection points. Are you retarded?

not true. Take f' = sinx + 2. There are infinite points of inflection for f but no roots of f' hence no local extrema for f.

>>8523605
Nigger, you should really think more carefully. The derivative is cos(x)

>>8523611
so what

>>8523605
nice

f(x)=x^3 is another counterexample

>>8523617
even OP's pic is a counter example if anyone looked at it

>>8523616
It also has infinite roots?

>>8523623
kek

you got me

I suppose you could add to your hypothesis that the function must have at least one maximum or minimum.

>>8523661
No, it should be that between n maximum and minimus there are n inflection points. Which it's a consecuence that between a max and min there is am inflection point.

>>8523641
That doesnt contradict anything I said

>>8523661
That would be equally wrong, look at the opening image

>>8523670
*n-1

>>8523675
Though OP is wrong. -sin(x) + 2x doesn't have inflection points because an inflection point must be an extremum, I.e. The derivative must be 0.

>>8523670
i'm pretty sure you can't get anything like this to work

for example, look at the portion of the graph in the OP between -2 and 2

if you imagine the graph is made of string nailed to the paper at these two points, you can create as many "hills" and "valleys" between these two points as you wan

you can do this without creating any new roots of f', but you can have arbitrarily many inflection points (which correspond to the peaks/troughs)

>>8523699
No it doesnt... first of all you integrated the function I described incorrectly, but even then 2x - sinx has infinite inflection points and no local extrema. An even simpler counter example is f = x^3, the derivative is 3x^2 which is nonnegative, so the function does not change from decreasing to increasing or vice versa anywhere, so has no local extrema. But f '' = 6x which clearly changes sign at x=0, thats its inflection point.

>>8523711
If you have a C^2 function, and the function has consecutive max min, there must be an inflection point. Just use the second derivative test and rolls theorem.

The actual result is that if there are n>0 local extrema then there are at least n-1 inflection points.

>>8523740
maybe my post wasn't clear, but i was talking about modifying the derivative, not the function itself

Thanks for your responses. I gather from this that i would have to amend "where 0 < n < infinity" to make this accurate.

I wonder if this has any application lol

>>8524003
Nigger read the thread again. You just have to change that it has at least n-1 points.

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>>8524015
After trying to write out a description with examples, i discovered that the theory does not hold up when any cubic function shows up.

Not sure if it's possible to fix this or if doing so would just lead us to Rolle's Theorem or the mean value theorem, or just a restatement of the definition of a derivative of f(x).

Oh well

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>>8524605
Sorry it's sideways, uploaded from my phone. Hopefully this one's better.

>>8524612
Perhaps it could function using critical values in place of mins/maxes...