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# Second derivative is 0 at a minimum point?

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1. What does it mean when the second derviative is 0 at a minimum or maximum point? Why isn't it positive or negative?
2. (Original post by Rana1997)
What does it mean when the second derviative is 0 at a minimum or maximum point? Why isn't it positive or negative?
Possibly a point of inflexion
3. (Original post by IYGB)
Possibly a point of inflexion
No it is not. Consider: y= x^4
4. (Original post by Rana1997)
No it is not. Consider: y= x^4
I never said it was. I said possibly
5. (Original post by IYGB)
I never said it was. I said possibly
All right. But what does that mean? Why is the gradient not changing while it is supposed to be?
6. (Original post by Rana1997)
All right. But what does that mean? Why is the gradient not changing while it is supposed to be?
If I remember well

Second derivative = zero AND third derivative = non zero, implies a point of inflexion.

Second derivative = zero AND third derivative = zero, implies the second derivative test fails and a different method must be used.
7. (Original post by Rana1997)
All right. But what does that mean? Why is the gradient not changing while it is supposed to be?
It just means that when the line is at x=0 then the line is momentarily flat.
8. (Original post by YouHaveProblems)
It just means that when the line is at x=0 then the line is momentarily flat.
That is when the first derivative is 0. The second derivitive is the change in the gradient which also becomes 0. This will make sense at an inflection point but it is illogical for minima and maxima.
9. (Original post by IYGB)
If I remember well

Second derivative = zero AND third derivative = non zero, implies a point of inflexion.

Second derivative = zero AND third derivative = zero, implies the second derivative test fails and a different method must be used.
Well, even in the first case the "second derivative test" has failed, since you are needing to look at the 3rd derivative as well.
If 2nd and 3rd. More generally, if the 1st, 2nd, ..., nth derivatives are zero but the n+1st is not, then if n is even we have a point of inflexion, if n is odd we have a maximum if the n+1st derivative is -ve, a minimum otherwise.

Note, however that it is possible for a (real) function to be infinitely differentiable and have a maximum point where all derivatives are zero.

(Original post by Rana1997)
That is when the first derivative is 0. The second derivitive is the change in the gradient which also becomes 0. This will make sense at an inflection point but it is illogical for minima and maxima.
Slight correction: the second derivative is the rate of change in the gradient. It's perfectly possible for the rate of change in the gradient to be zero at a maxima / minima. [Compare with the fact that the first derivative can be zero at a point without the function having to be constant there].

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