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derivatives and inflection points Watch

1. if the third derivative at a point does not equal zero and the second derivative equals zero, is that enough to guarantee the point is an inflection point?
2. (Original post by MEPS1996)
if the third derivative at a point does not equal zero and the second derivative equals zero, is that enough to guarantee the point is an inflection point?
Yes, they are sufficient conditions.

Note: Not all inflection points will have a third derivative non-zero, e.g. y=x^5 at x=0, where there is a point of inflection.
3. To prove inflection points I prefer the table method, as this takes longer time.

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4. (Original post by ghostwalker)
Yes, they are sufficient conditions.

Note: Not all inflection points will have a third derivative non-zero, e.g. y=x^5 at x=0, where there is a point of inflection.
so these are not sufficient conditions for all inflections points, but if these conditions are satisfied, then we have an inflection point. in practise if we had a function such as f(x)=x^5 then we could find that f''(0)=0, then we would see that f'''(0)=0, and conclude the test is inconclusive and just test nearby points. (not trying to get into more complex things). this is also the case with the second derivative test for finding out whether a point on y=f(x) is maxima or minima, by finding where f'(x)=0, then finding sign of f''(x). if 0, test is inconclusive. this is exactly the same as finding inflection points except we are finding maxima and minima of f'(x). i would be interested to know what is going on at minima and maxima which have second derivative equal to zero? can you offer any explanation as to why the gradient stops decreasing/ increasing?
thanks
5. (Original post by MEPS1996)
so these are not sufficient conditions for all inflections points, but if these conditions are satisfied, then we have an inflection point. in practise if we had a function such as f(x)=x^5 then we could find that f''(0)=0, then we would see that f'''(0)=0, and conclude the test is inconclusive and just test nearby points. (not trying to get into more complex things). this is also the case with the second derivative test for finding out whether a point on y=f(x) is maxima or minima, by finding where f'(x)=0, then finding sign of f''(x). if 0, test is inconclusive. this is exactly the same as finding inflection points except we are finding maxima and minima of f'(x). i would be interested to know what is going on at minima and maxima which have second derivative equal to zero? can you offer any explanation as to why the gradient stops decreasing/ increasing?
thanks
Nothing's "going on" - you just have to treat each case on its merits!

Note that if the 2nd derivative is 0 and the 3rd derivative isn't 0, then you do have an inflexion point, but note:

setting the second derivative to 0 won't find all inflexion points - the second derivative may not even exist at a point of inflexion!
Also, if the function is complicated then working out the 3rd derivative may be more trouble than examining the sign of the second derivative either side of the point!

The only guarantee you have is to work from the fundamental definition of an inflexion point: the curvature changes sign either side of that point.

Have you read about Taylor series? The Taylor series for a function about a point gives you a way of looking at its behaviour at a point where its first few derivatives vanish - the first non-vanishing derivative gives you a term like k(x-a)^n, and the parity (evenness or oddness) of n shows how it behaves.
6. a lot of the time, for the work involved, it`s far easier (for my tastes) just to work out the sign of the 2nd derivative on either side of the given point.

Another alternative is the "nth" derivative test.

say you have does that have a point of inflection at (0,0)??

1st, 2nd,3rd derivatives are all zero at x=0

But the 4th derivative is 72.

if, for a polynomial, for some n which is odd, the (n+1)th derivative at that point is >0 (which it is here), then that point is a local minimum. (<0 it`s a local max)

if you`re just dealing with then the 5th derivative of this {(4+1)=5} indicates (n is even=4) that (0,0) is an increasing inflection point.
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8. (Original post by davros)
Nothing's "going on" - you just have to treat each case on its merits!

Note that if the 2nd derivative is 0 and the 3rd derivative isn't 0, then you do have an inflexion point, but note:

setting the second derivative to 0 won't find all inflexion points - the second derivative may not even exist at a point of inflexion!
Also, if the function is complicated then working out the 3rd derivative may be more trouble than examining the sign of the second derivative either side of the point!

The only guarantee you have is to work from the fundamental definition of an inflexion point: the curvature changes sign either side of that point.

Have you read about Taylor series? The Taylor series for a function about a point gives you a way of looking at its behaviour at a point where its first few derivatives vanish - the first non-vanishing derivative gives you a term like k(x-a)^n, and the parity (evenness or oddness) of n shows how it behaves.
thanks so a second derivative which is discontinuos at a point where the limit as x goes to the x value of that point =0 could be an inflection point. or a second derivative which just goes from positve to negative without even going to zero. these could all be inflection points?
9. Inflection points are points in a curve where the gradient of the tangent to the curve is the maximum or minimum around the neighbourhood.
If you plot the gradient function (first derivative function) you'll see that the point corresponds to a maximum or minimum in this graph ( its mostly has a y value of 0).

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10. (Original post by MEPS1996)
thanks so a second derivative which is discontinuos at a point where the limit as x goes to the x value of that point =0 could be an inflection point. or a second derivative which just goes from positve to negative without even going to zero. these could all be inflection points?
The crucial point is that the curvature changes sign across an inflexion point, and hence the second derivative changes sign too.

If the second derivative exists there and is continuous, then we must have second derivative = 0, which is the "necessary" condition quoted by most text books for inflexion.

However, consider a function such as - here the point (0, 0) is an inflexion point because the curvature change sign, but the second derivative is not defined at that point

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